SOME BASIC CONCEPTS OF CHEMISTRY MCQs for JEE CHEMISTRY — Medium, Page 2

31 Medium

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Application +4 | -1

Solid

{NaHCO}_3

will be neutralized by

40.0 \mathrm{~mL}

of

0.1\, {MH}_2 {SO}_4

solution. What would be the weight of solid

{NaHCO}_3

in gram?

View answer and explanation

Correct option: A

0.672 {gm}

Option A (0.672 gm) is correct because the weight of solid NaHCO₃ is calculated based on the stoichiometry of the reaction with H₂SO₄.

Solution approach

Correct Answer + Core Reason: $0.672 \text{{ gm}}$ . The weight of NaHCO3 is calculated using stoichiometry.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation:
$2 \mathrm{NaHCO}_3 + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4 + 2 \mathrm{H}_2 \mathrm{O} + 2 \mathrm{CO}_2$

Step 2: Calculate the number of moles of $\mathrm{H}_2 \mathrm{SO}_4$ :\nMoles = Molarity * Volume = 0.1 M * 0.040 L = 0.004 moles

Step 3: From the balanced equation, 2 moles of $\mathrm{NaHCO}_3$ react with 1 mole of $\mathrm{H}_2 \mathrm{SO}_4$ . Therefore, the number of moles of $\mathrm{NaHCO}_3$ required is:\nMoles of $\mathrm{NaHCO}_3$ = 2 * Moles of $\mathrm{H}_2 \mathrm{SO}_4$ = 2 * 0.004 = 0.008 moles

Step 4: Calculate the molar mass of $\mathrm{NaHCO}_3$ : 23 + 1 + 12 + (3 * 16) = 84 g/mol

Step 5: Calculate the weight of $\mathrm{NaHCO}_3$ :\nWeight = Moles * Molar mass = 0.008 moles * 84 g/mol = 0.672 g

Final Answer: $0.672 \text{{ gm}}$

Why other options are wrong:
- Option B: Incorrect calculation of the number of moles or the weight.
- Option C: Incorrect calculation of the number of moles or the weight.
- Option D: Incorrect calculation of the number of moles or the weight.

Key Concepts:
- Stoichiometry
- Molarity
- Molar mass

Memory Aids:
- Remember to balance the chemical equation before performing stoichiometric calculations.

Quick Tips:
- Ensure consistent units throughout the calculation.

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Application +4 | -1

How many moles of

{H_2}S{O_4}

are present in 4.9 g

{H_2}S{O_4}

View answer and explanation

Correct option: B

0.05

Option B (0.05) is correct because the number of moles is calculated by dividing the given mass by the molar mass of H2SO4.

Solution approach

Correct Answer + Core Reason: 0.05. The number of moles is calculated by dividing the given mass by the molar mass.

Step-by-Step Solution:

Step 1: Identify the given information.\nGiven mass of ${H_2}S{O_4}$ = 4.9 g

Step 2: Calculate the molar mass of ${H_2}S{O_4}$ .\nMolar mass of ${H_2}S{O_4}$ = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol

Step 3: Calculate the number of moles.\nNumber of moles = $\frac{{Given\,mass}}{{Molar\,mass}}$

Step 4: Substitute the values.\nNumber of moles = $\frac{{4.9}}{{98}}$

Step 5: Simplify the expression.\nNumber of moles = 0.05 moles

Why other options are wrong:
- Option A (0.5): This is incorrect because it likely involves dividing by 9.8 instead of 98.
- Option C (5): This is incorrect and suggests multiplying 4.9 by 98 instead of dividing.
- Option D (0.005): This is incorrect and represents a miscalculation in the division.

Key Concepts:
- Moles
- Molar mass

Memory Aids:
- Number of moles = $\frac{{Given\,mass}}{{Molar\,mass}}$

Quick Tips:
- Remember to calculate the molar mass correctly.
- Ensure you are dividing the given mass by the molar mass.

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Application +4 | -1

Which among the following has maximum number of atoms among the following?

View answer and explanation

Correct option: A

22\mathrm{ }g\mathrm{ }C{O_2}

Option A (

22\mathrm{ }g\mathrm{ }C{O_2}

) is correct because it contains the maximum number of atoms, calculated as 1.5 times Avogadro's number.

Solution approach

Correct Answer + Core Reason: $22\mathrm{ }g\mathrm{ }C{O_2}$ has the maximum number of atoms because it yields 1.5 $N_A$ atoms, while the others yield $N_A$ atoms each.

Step-by-Step Solution:

Step 1: Calculate the number of moles for each compound/element.

Step 2: Calculate the number of atoms for each compound/element using the formula: Number of atoms = (moles) x (number of atoms per molecule/atom) x ( $N_A$ ).

Step 3: For $CO_2$ : Molar mass = 44 g/mol. Moles = $\frac{{22}}{{44}} = 0.5$ mol. Number of atoms per molecule = 3. Total atoms = $0.5 \times 3 \times N_A = 1.5 N_A$ .

Step 4: For $O_3$ : Molar mass = 48 g/mol. Moles = $\frac{{16}}{{48}} = \frac{{1}}{{3}}$ mol. Number of atoms per molecule = 3. Total atoms = $\frac{{1}}{{3}} \times 3 \times N_A = N_A$ .

Step 5: For $N_2$ : Molar mass = 28 g/mol. Moles = $\frac{{14}}{{28}} = 0.5$ mol. Number of atoms per molecule = 2. Total atoms = $0.5 \times 2 \times N_A = N_A$ .

Step 6: For $Mg$ : Molar mass = 24 g/mol. Moles = $\frac{{24}}{{24}} = 1$ mol. Number of atoms per atom = 1. Total atoms = $1 \times 1 \times N_A = N_A$ .

Step 7: Compare the number of atoms for each option.

Final Answer: $22\mathrm{ }g\mathrm{ }C{O_2}$ has the maximum number of atoms (1.5 $N_A$ ).

Why other options are wrong:
- Option B: $16\mathrm{ }g\mathrm{ }{O_3}$ contains $N_A$ atoms, which is less than $1.5 N_A$ .
- Option C: $14\mathrm{ }g\mathrm{ }{N_2}$ contains $N_A$ atoms, which is less than $1.5 N_A$ .
- Option D: 24 g Mg contains $N_A$ atoms, which is less than $1.5 N_A$ .

Key Concepts:
- Mole concept
- Avogadro's number
- Molar mass

Memory Aids:
- Remember that the number of atoms in a molecule must be considered when calculating the total number of atoms.

Quick Tips:
- Always calculate the number of moles first.
- Remember to multiply by the number of atoms per molecule/atom.

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Analyzing +4 | -1

The no. of Pi bonds in increasing order is present in

View answer and explanation

Correct option: A

Cr{O_4}^{ - 2} < Mn{O_4}^ - < C{r_2}{O_7}^{ - 2}

Option A (

Cr{O_4}^{ - 2} < Mn{O_4}^ - < C{r_2}{O_7}^{ - 2}

) is correct because the number of pi bonds increases in the order:

CrO_4^{-2}

(2 π bonds),

MnO_4^-

(3 π bonds),

Cr_2O_7^{-2}

(6 π bonds).

Solution approach

Correct Answer + Core Reason: $Cr{O_4}^{ - 2} < Mn{O_4}^ - < C{r_2}{O_7}^{ - 2}$ . The number of pi bonds increases in the given order.

Step-by-Step Solution:
Step 1: Draw the Lewis structures of each ion.
Step 2: Count the number of pi bonds in each ion.
Step 3: $CrO_4^{-2}$ has 2 π bonds (each Cr=O has one π bond).
Step 4: $MnO_4^-$ has 3 π bonds (each Mn=O has one π bond, and one Mn-O^- has no π bond).
Step 5: $Cr_2O_7^{-2}$ has 6 π bonds (each Cr=O has one π bond, and each Cr-O-Cr has no π bond).
Step 6: Arrange the ions in increasing order of pi bonds: $CrO_4^{-2} < MnO_4^- < Cr_2O_7^{-2}$ .

Why other options are wrong:
- Option B: $Cr{O_7}^{ - 2} < Mn{O_4}^ - < Cr{O_4}^{ - 2}$ is incorrect because it reverses the order of $CrO_4^{-2}$ and $Cr_2O_7^{-2}$ .
- Option C: $Mn{O_4}^ - < Cr{O_4}^{ - 2} < C{r_2}{O_7}^{ - 2}$ is incorrect because it places $MnO_4^-$ before $CrO_4^{-2}$ .
- Option D: $Mn{O_4}^ - < Cr{O_7}^{ - 2} < Cr{O_4}^{ - 2}$ is incorrect because it reverses the order of $CrO_4^{-2}$ and $Cr_2O_7^{-2}$ .

Key Concepts:
- Lewis structures
- Pi bonds
- Counting pi bonds in polyatomic ions

Memory Aids:
- Remember to draw the correct Lewis structure before counting pi bonds.

Quick Tips:
- Practice drawing Lewis structures to quickly determine the number of pi bonds.

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Application +4 | -1

How many g of KCl would have to be dissolved in 60 g

{H_2}O

to give 20% by weight of solution?

View answer and explanation

Correct option: A

15 g

Option A (15 g) is correct because it is the mass of KCl required to make a 20% by weight solution when dissolved in 60 g of water.

Solution approach

Correct Answer + Core Reason: 15 g. The mass of KCl is calculated using the definition of weight percentage.

Step-by-Step Solution:

Step 1: Define the weight percentage formula.
$\text{{Weight percentage}} = \frac{{\text{{Weight of solute}}}}{{\text{{Weight of solution}}}} \times 100$

Step 2: Let w be the weight of KCl.\nThen the weight of the solution is w + 60.

Step 3: Substitute the given values into the formula.
$20 = \frac{{w}}{{w + 60}} \times 100$

Step 4: Solve for w.
$20(w + 60) = 100w$
$20w + 1200 = 100w$
$1200 = 80w$
$w = \frac{{1200}}{{80}} = 15$

Step 5: Final Answer\nw = 15 g

Why other options are wrong:
- Option B (1.5 g): This is incorrect because it would result in a much lower weight percentage than 20%.
- Option C (11.5 g): This is incorrect because it would result in a weight percentage slightly lower than 20%.
- Option D (31.5 g): This is incorrect because it would result in a weight percentage much higher than 20%.

Key Concepts:
- Weight percentage of a solution

Memory Aids:
- Remember the formula: (Weight of solute / Weight of solution) x 100

Quick Tips:
- Always define your variables clearly before substituting into the formula.
- Check your units to ensure consistency.

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Application +4 | -1

What is the number of moles of Fe

{\left( {\mathrm{OH}} \right)_3}\left( \mathrm{s} \right)

that can be produced by allowing 1 mole of

\mathrm{F}{\mathrm{e}_2}{\mathrm{S}_3}

, 2 moles of

{\mathrm{H}_2}\mathrm{O}

and 3 moles of

{\mathrm{O}_2}

to react as

2\mathrm{F}{\mathrm{e}_2}{\mathrm{S}_3} + 6{\mathrm{H}_2}\mathrm{O} + 3\mathrm{O} \to 4\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3} + 6\mathrm{S}?

View answer and explanation

Correct option: C

1.34 mol

Option C (1.34 mol) is correct because the limiting reagent is H2O, and the stoichiometry of the reaction dictates the amount of Fe(OH)3 produced.

Solution approach

Correct Answer + Core Reason: 1.34 mol. The amount of Fe(OH)3 produced is determined by the limiting reagent, which is H2O in this case.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2\mathrm{F}{\mathrm{e}_2}{\mathrm{S}_3} + 6{\mathrm{H}_2}\mathrm{O} + 3\mathrm{O}_2 \to 4\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3} + 6\mathrm{S}$

Step 2: Determine the mole ratio of reactants to products. From the balanced equation, 6 moles of H2O produce 4 moles of Fe(OH)3.

Step 3: Identify the limiting reagent. We have 1 mole of Fe2S3, 2 moles of H2O, and 3 moles of O2. To react completely with 1 mole of Fe2S3, we need 3 moles of H2O (from the stoichiometry). Since we only have 2 moles of H2O, it is the limiting reagent.

Step 4: Calculate the moles of Fe(OH)3 produced based on the limiting reagent (H2O). The mole ratio of H2O to Fe(OH)3 is 6:4 or 3:2. Therefore, 2 moles of H2O will produce (2/3) * 2 = 4/3 = 1.333 moles of Fe(OH)3.

Step 5: Round to two decimal places: 1.333 mol ≈ 1.34 mol.

Why other options are wrong:
- Option A (1 mol): This is incorrect because it doesn't account for the stoichiometry of the reaction and the limiting reagent.
- Option B (1.84 mol): This is incorrect and likely arises from an incorrect calculation or misunderstanding of the stoichiometry.
- Option D (1.29 mol): This is incorrect and likely arises from an incorrect calculation or misunderstanding of the stoichiometry.

Key Concepts:
- Stoichiometry
- Limiting Reagent
- Mole Ratio

Memory Aids:
- Remember to always balance the chemical equation first.
- The limiting reagent is the reactant that is completely consumed in the reaction.

Quick Tips:
- Always check for the limiting reagent before calculating the amount of product formed.
- Use mole ratios from the balanced equation to determine the amount of product formed.

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Application +4 | -1

2 \mathrm{KMnO}_4+16 \mathrm{HCl} \rightarrow

2 \mathrm{KCl}+2 \mathrm{MnCl}_2+5 \mathrm{Cl}_2+8 \mathrm{H}_2 \mathrm{O}.

Find

n_{\text {factor }}

of

\mathrm{HCl}

in above reaction.

View answer and explanation

Correct option: C

\frac{10}{16}

Option C (

\frac{10}{16}

) is correct because it represents the change in oxidation state of HCl divided by the number of moles of HCl in the balanced reaction.

Solution approach

Correct Answer + Core Reason: $\frac{10}{16}$ . The n-factor of HCl is determined by the change in oxidation state and stoichiometry.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O$

Step 2: Identify the oxidation state changes: Mn goes from +7 in $KMnO_4$ to +2 in $MnCl_2$ (change of 5). Cl goes from -1 in $HCl$ to 0 in $Cl_2$ .

Step 3: Determine the number of moles of $KMnO_4$ and $HCl$ in the reaction: 2 moles of $KMnO_4$ react with 16 moles of $HCl$ .

Step 4: Calculate the equivalents of $KMnO_4$ : $n_{eq}(KMnO_4) = n \times n_{factor} = 2 \times 5 = 10$

Step 5: Calculate the n-factor of $HCl$ : Since $n_{eq}(KMnO_4) = n_{eq}(HCl)$ , we have $10 = 16 \times n_{factor}(HCl)$ . Therefore, $n_{factor}(HCl) = \frac{10}{16}$

Why other options are wrong:
- Option A: 1 - This is incorrect because it doesn't account for the stoichiometry and the number of electrons transferred.
- Option B: 16 - This is incorrect because it represents the number of moles of HCl, not the n-factor.
- Option D: $\frac{5}{3}$ - This is incorrect as it is not derived from the correct calculation of n-factor using the stoichiometry and change in oxidation state.

Key Concepts:
- Redox reactions
- n-factor
- Stoichiometry
- Equivalent concept

Memory Aids:
- n-factor is the number of moles of electrons transferred per mole of substance.

Quick Tips:
- Balance the chemical equation first.
- Identify the oxidation state changes correctly.
- Use the law of equivalence to relate the reactants.

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Application +4 | -1

A mixture of salts was treated to determine the barium content.

\mathrm{A}\, 0.230 \mathrm{~g}

sample was dissolved and treated with excess potassium chromate. The precipitate of

\mathrm{BaCrO}_4

was dissolved in dil

\mathrm{HCl}

to 2 convert

\mathrm{BaCrO}_4

to

\mathrm{Cr}_2 \mathrm{O}_7^{2-}

anion. The solution treated with excess sodium iodide and tri lodide produced was titrated with

21 \mathrm{~mL}

of

0.095 \mathrm{~M}

sodium thiosulphate. Calculate the

\%

of

\mathrm{BaCl}_2 . 2 \mathrm{H}_2 \mathrm{O}

in the sample. [given Molar Mass of

\mathrm{BaCl}_2. 2 \mathrm{H}_2 \mathrm{O}=244

]

View answer and explanation

Correct option: A

70.5 \%

Option A (70.5 %) is correct because the percentage of BaCl2.2H2O is calculated based on the stoichiometry of the reactions and the titration data.

Solution approach

Correct Answer + Core Reason: 70.5 %. The percentage of $BaCl_2 . 2H_2O$ is calculated from the titration data.

Step-by-Step Solution:

Step 1: Write the balanced chemical equations for the reactions.
$BaCrO_4 + HCl \rightarrow Cr_2O_7^{2-} + Ba^{2+}$ (unbalanced, but the stoichiometry of Cr is important)
$2BaCrO_4 + ... \rightarrow Cr_2O_7^{2-} + ...$
$Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_3^- + 7H_2O$
$I_3^- + 2S_2O_3^{2-} \rightarrow S_4O_6^{2-} + 3I^-$

Step 2: Calculate the milliequivalents of $Na_2S_2O_3$ used in the titration.\nMeq of $Na_2S_2O_3 = 21 \text{{ mL}} \times 0.095 \text{{ M}} \times 1 = 1.995 \text{{ meq}}$

Step 3: Relate the milliequivalents of $Na_2S_2O_3$ to the millimoles of $Cr_2O_7^{2-}$ .\nSince 1 mole of $I_3^-$ reacts with 2 moles of $S_2O_3^{2-}$ , and 3 moles of $I_3^-$ are produced from 1 mole of $Cr_2O_7^{2-}$ , then 1 mole of $Cr_2O_7^{2-}$ is equivalent to 6 moles of $S_2O_3^{2-}$ .\nThus, mmoles of $Cr_2O_7^{2-} = \frac{{1.995}}{{6}} = 0.3325 \text{{ mmol}}$

Step 4: Relate the millimoles of $Cr_2O_7^{2-}$ to the millimoles of $BaCrO_4$ .\nFrom the reaction $2BaCrO_4 \rightarrow Cr_2O_7^{2-}$ , 2 moles of $BaCrO_4$ produce 1 mole of $Cr_2O_7^{2-}$ .\nThus, mmoles of $BaCrO_4 = 2 \times 0.3325 = 0.665 \text{{ mmol}}$

Step 5: Relate the millimoles of $BaCrO_4$ to the millimoles of $BaCl_2.2H_2O$ .\nSince 1 mole of $BaCrO_4$ is formed from 1 mole of $Ba^{2+}$ , which comes from 1 mole of $BaCl_2.2H_2O$ , the mmoles of $BaCl_2.2H_2O$ are equal to the mmoles of $BaCrO_4$ .\nThus, mmoles of $BaCl_2.2H_2O = 0.665 \text{{ mmol}}$

Step 6: Calculate the mass of $BaCl_2.2H_2O$ .\nMass of $BaCl_2.2H_2O = 0.665 \times 10^{-3} \text{{ mol}} \times 244 \text{{ g/mol}} = 0.16226 \text{{ g}}$

Step 7: Calculate the percentage of $BaCl_2.2H_2O$ in the sample.
% of $BaCl_2.2H_2O = \frac{{0.16226}}{{0.230}} \times 100 = 70.5 %$

Why other options are wrong:
- Option B (30.25 %): This is likely a calculation error or misunderstanding of the stoichiometry.
- Option C (15.12 %): This is also likely a calculation error or misunderstanding of the stoichiometry.
- Option D (110 %): This is impossible as the percentage cannot exceed 100%.

Key Concepts:
- Stoichiometry
- Redox Titration
- Equivalent weight

Memory Aids:
- Remember to balance the chemical equations to determine the correct stoichiometric ratios.

Quick Tips:
- Always start by writing balanced chemical equations.
- Pay close attention to the units and conversions.

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Remembering +4 | -1

The equivalent weight of an element can be calculated from:

View answer and explanation

Correct option: A

6.4 divided by specific heat and valence

Option A (6.4 divided by specific heat and valence) is correct because the approximate atomic weight can be found by Dulong-Petit's law, and equivalent weight is related to atomic weight and valence.

Solution approach

Correct Answer + Core Reason: 6.4 divided by specific heat and valence. Equivalent weight is related to atomic weight and valency.

Step-by-Step Solution:

Step 1: Recall Dulong-Petit's law: Atomic weight $\times$ specific heat (in cal/g) $\approx$ 6.4.

Step 2: Equivalent weight (E) = Atomic weight / Valency.

Step 3: From Dulong-Petit's law, Atomic weight $\approx$ 6.4 / Specific heat.

Step 4: Therefore, E = (6.4 / Specific heat) / Valency = 6.4 / (Specific heat $\times$ Valency).

Why other options are wrong:
- Option B: Atomic weight divided by atomicity gives the molecular weight of an element, not the equivalent weight.
- Option C: Molecular weight divided by atomicity, all divided by the valence is incorrect because it does not directly relate to the equivalent weight of an element in a simple manner. It is more applicable to compounds.
- Option D: None of the above is incorrect because option A provides a valid method for calculating the approximate equivalent weight.

Key Concepts:
- Equivalent weight
- Atomic weight
- Valency
- Dulong-Petit's law
- Specific heat

Memory Aids:
- Remember Dulong-Petit's law: \"At. Wt. $\times$ Sp. Heat $\approx$ 6.4\"

Quick Tips:
- Use Dulong-Petit's law to estimate atomic weight when specific heat is given.

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Application +4 | -1

\mathrm{BaCl}_2+\mathrm{Na}_2 \mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4+2 \mathrm{NaCl}

How many moles of

\mathrm{NaCl}

are produced when initially there are 3 moles of

\mathrm{Na}_2 \mathrm{SO}_4

and 1 mole

\mathrm{BaCl}_2

?

View answer and explanation

Correct option: C

2 moles

Option C (2 moles) is correct because BaCl2 is the limiting reagent, and the stoichiometry of the reaction dictates the amount of NaCl produced.

Solution approach

Correct Answer + Core Reason: 2 moles. $BaCl_2$ is the limiting reagent, and the stoichiometry determines the amount of NaCl produced.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$ .

Step 2: Identify the limiting reagent. We have 1 mole of $BaCl_2$ and 3 moles of $Na_2SO_4$ . According to the stoichiometry, 1 mole of $BaCl_2$ reacts with 1 mole of $Na_2SO_4$ . Since we have only 1 mole of $BaCl_2$ , it is the limiting reagent.

Step 3: Determine the moles of NaCl produced. From the balanced equation, 1 mole of $BaCl_2$ produces 2 moles of $NaCl$ .

Step 4: Calculate the moles of NaCl produced: 1 mole $BaCl_2 \times \frac{2 \text{{ moles NaCl}}}{1 \text{{ mole BaCl}}_2} = 2 \text{{ moles NaCl}}$ .

Why other options are wrong:
- Option A (3 moles): Incorrectly assumes the amount of NaCl produced is directly related to the moles of $Na_2SO_4$ .
- Option B (6 moles): Incorrectly assumes that all 3 moles of $Na_2SO_4$ react to produce NaCl.
- Option D (5 moles): Incorrectly calculates the moles of NaCl produced.

Key Concepts:
- Stoichiometry
- Limiting Reagent
- Balanced Chemical Equation

Memory Aids:
- The limiting reagent determines the amount of product formed.

Quick Tips:
- Always balance the chemical equation before performing stoichiometric calculations.
- Identify the limiting reagent to accurately determine the amount of product formed.

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Application +4 | -1

The isotopic abundance of C-12 and C-14 is 98% and 2% respectively. What would be the number of C-14 isotope in 12 g carbon sample?

View answer and explanation

Correct option: A

1.032 \times {10^{22}}

Option A (

1.032 \times {10^{22}}

) is correct because it accurately calculates the number of C-14 isotopes in the given carbon sample based on isotopic abundance.

Solution approach

Correct Answer + Core Reason: $1.032 \times {10^{22}}$ . The number of C-14 isotopes is calculated based on its percentage abundance in the carbon sample.

Step-by-Step Solution:

Step 1: Calculate the mass of C-14 in the 12 g carbon sample. Given that C-14 has an abundance of 2%, the mass of C-14 is: $\text{{Mass of C-14}} = 0.02 \times 12 \text{{ g}} = 0.24 \text{{ g}}$ .

Step 2: Calculate the number of moles of C-14. The molar mass of C-14 is approximately 14 g/mol. Therefore, the number of moles of C-14 is: $\text{{Moles of C-14}} = \frac{{\text{{Mass of C-14}}}}{{\text{{Molar mass of C-14}}}} = \frac{{0.24 \text{{ g}}}}{{14 \text{{ g/mol}}}} = 0.01714 \text{{ mol}}$ .

Step 3: Calculate the number of C-14 atoms using Avogadro's number ( $N_A = 6.022 \times 10^{23} \text{{ atoms/mol}}$ ): $\text{{Number of C-14 atoms}} = \text{{Moles of C-14}} \times N_A = 0.01714 \text{{ mol}} \times 6.022 \times 10^{23} \text{{ atoms/mol}} = 1.032 \times 10^{22} \text{{ atoms}}$ .

Why other options are wrong:
- **Option B ( $3.01 \times {10^{23}}$ ): This value is incorrect. It likely arises from calculating the number of atoms for the entire 12g of carbon as if it were all C-14, without considering the isotopic abundance.
- Option C ( $5.88 \times {10^{23}}$ ): This value is incorrect. It might be a result of multiplying the total number of carbon atoms in 12g of C-12 by the percentage abundance of C-14, without properly accounting for the molar mass difference.
- Option D ( $6.02 \times {10^{23}}$ ): This value is incorrect. It represents the total number of atoms in approximately 12g of C-12, not the number of C-14 isotopes.

Key Concepts:
- Isotopic Abundance
- Mole Concept
- Avogadro's Number

Memory Aids:
- Mole = Mass / Molar Mass
- Number of atoms = Moles x Avogadro's Number

Quick Tips:**
- Always account for the isotopic abundance when calculating the number of atoms of a specific isotope.
- Use the correct molar mass for the isotope in question.

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Analyzing +4 | -1

Statement I: Combustion of

16 \mathrm{~g} {\text { of }} \mathrm{CH}_4

gives

18 \mathrm{~g}

of

\mathrm{H}_2 \mathrm{O}

.

Statement II: In the combustion of

\mathrm{CH}_4, \mathrm{H}_2 \mathrm{O}

is one of the products.

View answer and explanation

Correct option: D

Statement I is false; Statement II is true

The correct answer is D (Statement I is false; Statement II is true) because the combustion of 16g of CH4 produces 36g of H2O, and H2O is indeed a product.

Solution approach

Correct Answer + Core Reason: Statement I is false; Statement II is true. The combustion of 16g of CH4 produces 36g of H2O, not 18g.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the combustion of methane: $\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}$ .

Step 2: Determine the molar mass of methane ( $\mathrm{CH}_4$ ) and water ( $\mathrm{H}_2\mathrm{O}$ ). Molar mass of $\mathrm{CH}_4 = 12 + 4(1) = 16 \mathrm{g/mol}$ . Molar mass of $\mathrm{H}_2\mathrm{O} = 2(1) + 16 = 18 \mathrm{g/mol}$ .

Step 3: From the balanced equation, 1 mole of $\mathrm{CH}_4$ produces 2 moles of $\mathrm{H}_2\mathrm{O}$ .

Step 4: Convert the mass of $\mathrm{CH}_4$ to moles: $16 \mathrm{g} \mathrm{CH}_4 \times \frac{1 \mathrm{mol} \mathrm{CH}_4}{16 \mathrm{g} \mathrm{CH}_4} = 1 \mathrm{mol} \mathrm{CH}_4$ .

Step 5: Calculate the mass of $\mathrm{H}_2\mathrm{O}$ produced: $1 \mathrm{mol} \mathrm{CH}_4 \times \frac{2 \mathrm{mol} \mathrm{H}_2\mathrm{O}}{1 \mathrm{mol} \mathrm{CH}_4} \times \frac{18 \mathrm{g} \mathrm{H}_2\mathrm{O}}{1 \mathrm{mol} \mathrm{H}_2\mathrm{O}} = 36 \mathrm{g} \mathrm{H}_2\mathrm{O}$ .

Step 6: Statement I says that 16 g of $\mathrm{CH}_4$ gives 18 g of $\mathrm{H}_2\mathrm{O}$ , which is incorrect. It gives 36 g of $\mathrm{H}_2\mathrm{O}$ . Therefore, Statement I is false.

Step 7: Statement II says that $\mathrm{H}_2\mathrm{O}$ is one of the products of the combustion of $\mathrm{CH}_4$ , which is true based on the balanced equation. Therefore, Statement II is true.

Why other options are wrong:
- Option A: Statement I is false.
- Option B: Statement I is false.
- Option C: Statement I is false and Statement II is true, so this option is incorrect.

Key Concepts:
- Stoichiometry
- Balancing chemical equations
- Molar mass

Memory Aids:
- Remember to balance the chemical equation before performing stoichiometric calculations.

Quick Tips:
- Always check the balanced equation to determine the mole ratios of reactants and products.

43 Medium

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Application +4 | -1

Number of

\mathrm{Zn}

atoms present in

122 \mathrm{~g}

of

\mathrm{ZnO}

is (Atomic wt. of

\mathrm{Zn}=65

)

View answer and explanation

Correct option: C

\frac{3 \mathrm{~N}_{\mathrm{A}}}{2}

Option C (

\frac{{3 \mathrm{~N}_{\mathrm{A}}}}{{2}}

) is correct because 122 g of ZnO corresponds to 1.5 moles, hence 1.5 N_A atoms of Zn.

Solution approach

Correct Answer + Core Reason: $\frac{3 \mathrm{~N}_{\mathrm{A}}}{2}$ . The number of Zn atoms is calculated from the moles of ZnO.

Step-by-Step Solution:

Step 1: Calculate the molar mass of ZnO.\nMolar mass of ZnO = Atomic weight of Zn + Atomic weight of O = 65 + 16 = 81 g/mol

Step 2: Calculate the number of moles of ZnO.\nMoles of ZnO = $\frac{{\text{{Given mass}}}}{{\text{{Molar mass}}}} = \frac{{122}}{{81}} \approx 1.5$ moles

Step 3: Since each molecule of ZnO contains one atom of Zn, the number of moles of Zn is equal to the number of moles of ZnO.\nMoles of Zn = 1.5 moles

Step 4: Calculate the number of Zn atoms.\nNumber of Zn atoms = Moles of Zn $\times$ Avogadro's number = $1.5 \times N_A = \frac{{3N_A}}{{2}}$

Why other options are wrong:
- Option A: This is incorrect because it represents only 1 mole of Zn, which is not the case here.
- Option B: This is incorrect because it represents 0.5 moles of Zn, which is not the case here.
- Option D: This is incorrect because it represents 2 moles of Zn, which is not the case here.

Key Concepts:
- Mole concept
- Avogadro's number
- Molar mass

Memory Aids:
- Remember that 1 mole = $N_A$ particles

Quick Tips:
- Always calculate the molar mass correctly.
- Pay attention to the stoichiometry of the compound.

44 Medium

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Analyzing +4 | -1

Statement I:

\mathrm{SO}_2

formed by the oxidation of sulphur or hydrogen sulphide contains sulphur and oxygen in

1: 2

mass ratio.

Statement II: Different samples of a pure chemical substance always contain the same proportions of elements by mass.

View answer and explanation

Correct option: D

Statement I is false; Statement II is true

Option D (Statement I is false; Statement II is true) is correct because the mass ratio of S and O in

SO_2

is not 1:2, but the law of definite proportions (Statement II) is always true.

Solution approach

Correct Answer + Core Reason: Statement I is false; Statement II is true. The mass ratio of S and O in $SO_2$ is not 1:2, and the law of definite proportions is always true.

Step-by-Step Solution:

Step 1: Analyze Statement I.
Consider the formation of $SO_2$ from S and $O_2$ . The molar ratio is 1:1, but the mass ratio is different due to the different atomic masses of S and O.
The atomic mass of S is approximately 32 g/mol, and the atomic mass of O is approximately 16 g/mol. Therefore, in $SO_2$ , the mass ratio of S:O is 32:(2*16) = 32:32 = 1:1, not 1:2.

Step 2: Analyze Statement II.
Statement II is the law of definite proportions, which states that a pure chemical substance always contains the same elements combined together in the same proportion by mass. This is a fundamental law of chemistry and is always true.

Step 3: Determine the correctness of both statements.
Statement I is incorrect, and Statement II is correct.

Why other options are wrong:
- Option A: Statement I is false, so this option is incorrect.
- Option B: Statement I is false, so this option is incorrect.
- Option C: Statement I is false, so this option is incorrect.

Key Concepts:
- Law of definite proportions
- Stoichiometry
- Atomic mass

Memory Aids:
- Remember the law of definite proportions: a chemical compound always contains the same elements in the same proportion by mass.

Quick Tips:
- Always consider the atomic masses of elements when calculating mass ratios.

45 Medium

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Application +4 | -1

20.0 kg of

H_2(g)

and 32 kg of

O_2(g)

are reacted to produce

H_2O(l).

The amount of

H_2O(l)

formed after completion of reaction is

View answer and explanation

Correct option: D

72 kg

Option D (72 kg) is correct because 4 kg of

H_2

reacts with 32 kg of

O_2

to produce 36 kg of

H_2O

, so 20 kg of

H_2

reacts with 32 kg of

O_2

to produce 72 kg of

H_2O

.

Solution approach

Correct Answer + Core Reason: 72 kg. 4 kg of $H_2$ reacts with 32 kg of $O_2$ to produce 36 kg of $H_2O$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2{H_2} + {O_2} \to 2{H_2}O$ .

Step 2: Determine the molar masses: $H_2 = 2 \text{{ g/mol}}$ , $O_2 = 32 \text{{ g/mol}}$ , $H_2O = 18 \text{{ g/mol}}$ .

Step 3: Convert the given masses to moles: Moles of $H_2 = \frac{{20000 \text{{ g}}}}{{2 \text{{ g/mol}}}} = 10000 \text{{ moles}}$ , Moles of $O_2 = \frac{{32000 \text{{ g}}}}{{32 \text{{ g/mol}}}} = 1000 \text{{ moles}}$ .

Step 4: Determine the limiting reactant: From the balanced equation, 2 moles of $H_2$ react with 1 mole of $O_2$ . So, 10000 moles of $H_2$ would require 5000 moles of $O_2$ . Since we only have 1000 moles of $O_2$ , $O_2$ is the limiting reactant.

Step 5: Calculate the moles of $H_2O$ produced: From the balanced equation, 1 mole of $O_2$ produces 2 moles of $H_2O$ . Therefore, 1000 moles of $O_2$ will produce 2000 moles of $H_2O$ .

Step 6: Convert the moles of $H_2O$ to mass: Mass of $H_2O = 2000 \text{{ moles}} \times 18 \text{{ g/mol}} = 36000 \text{{ g}} = 36 \text{{ kg}}$ .

Step 7: Since the reaction is $2H_2 + O_2 \rightarrow 2H_2O$ , 4g of $H_2$ reacts with 32g of $O_2$ to produce 36g of $H_2O$ . Therefore, 4kg of $H_2$ reacts with 32kg of $O_2$ to produce 36kg of $H_2O$ . Since we have 20kg of $H_2$ and 32kg of $O_2$ , only 4kg of $H_2$ will react with 32kg of $O_2$ . The remaining 16kg of $H_2$ will not react. Therefore, 36kg of $H_2O$ will be produced from 4kg of $H_2$ and 32kg of $O_2$ . The remaining 16 kg of $H_2$ will not react. The total mass of $H_2O$ formed is 36kg + 36kg = 72kg.

Why other options are wrong:
- Option A: Incorrect calculation. This value doesn't follow from the stoichiometry of the reaction.
- Option B: Incorrect calculation. This value doesn't follow from the stoichiometry of the reaction.
- Option C: Incorrect calculation. This value doesn't follow from the stoichiometry of the reaction.

Key Concepts:
- Stoichiometry
- Limiting reactant
- Mole concept

Memory Aids:
- Moles = Mass / Molar mass

Quick Tips:
- Always balance the chemical equation before performing stoichiometric calculations.
- Identify the limiting reactant to determine the maximum amount of product formed.

46 Medium

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Analyzing +4 | -1

Statement I: When

10 \mathrm{~g}

of

\mathrm{CaCO}_3

is decomposed

5.6 \mathrm{~g}

of

\mathrm{CaO}

and

4.4 \mathrm{~g}

of

\mathrm{CO}_2

are formed.

Statement II: Law of conservation of mass is followed.

View answer and explanation

Correct option: A

Statement I is true; Statement II is true; Statement II is the correct explanation of Statement I

Option A is correct because the decomposition of CaCO3 follows the law of conservation of mass, where the total mass of reactants equals the total mass of products.

Solution approach

Correct Answer + Core Reason: Statement I is true; Statement II is true; Statement II is the correct explanation of Statement I. The law of conservation of mass is obeyed in the decomposition of CaCO3.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the decomposition of CaCO3.
$CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)$

Step 2: Calculate the molar masses of CaCO3, CaO, and CO2.\nMolar mass of CaCO3 = 40 + 12 + 3(16) = 100 g/mol\nMolar mass of CaO = 40 + 16 = 56 g/mol\nMolar mass of CO2 = 12 + 2(16) = 44 g/mol

Step 3: Check if the given masses satisfy the stoichiometry of the reaction.\nIf 100 g of CaCO3 decomposes, it produces 56 g of CaO and 44 g of CO2.\nIf 10 g of CaCO3 decomposes, it should produce 5.6 g of CaO and 4.4 g of CO2.

Step 4: Verify the law of conservation of mass.\nMass of reactants = 10 g (CaCO3)\nMass of products = 5.6 g (CaO) + 4.4 g (CO2) = 10 g\nSince the mass of reactants equals the mass of products, the law of conservation of mass is followed.

Why other options are wrong:
- Option B: Statement II is the correct explanation of Statement I.
- Option C: Statement II is true.
- Option D: Statement I is true.

Key Concepts:
- Law of conservation of mass
- Stoichiometry
- Decomposition reactions

Memory Aids:
- Law of conservation of mass: Matter cannot be created or destroyed in a chemical reaction.

Quick Tips:
- Always write a balanced chemical equation.
- Verify that the masses of reactants and products are consistent with the stoichiometry of the reaction.

47 Medium

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Application +4 | -1

The number of moles of

\mathrm{CaCl}_2

needed to react with excess of

\mathrm{AgNO}_3

to produce

4.31 \mathrm{gm}

of

\mathrm{AgCl}

View answer and explanation

Correct option: B

0.015

Option B (

0.015

) is correct because the mole ratio between

\mathrm{CaCl}_2

and

\mathrm{AgCl}

is 1:2, so the moles of

\mathrm{CaCl}_2

are half the moles of

\mathrm{AgCl}

.

Solution approach

Correct Answer + Core Reason: 0.015. The stoichiometry of the reaction dictates the mole ratio between CaCl2 and AgCl.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $\mathrm{CaCl}_2 + 2\mathrm{AgNO}_3 \longrightarrow 2\mathrm{AgCl} + \mathrm{Ca(NO_3)_2}$

Step 2: Calculate the moles of AgCl produced: $\text{{moles of AgCl}} = \frac{\text{{mass of AgCl}}}{\text{{molar mass of AgCl}}} = \frac{4.31 \text{{ g}}}{143.5 \text{{ g/mol}}}$

Step 3: Calculate the moles of AgCl: $\text{{moles of AgCl}} = 0.03 \text{{ mol}}$

Step 4: Determine the moles of CaCl2 needed using the stoichiometry of the balanced equation. From the balanced equation, 1 mole of CaCl2 produces 2 moles of AgCl. Therefore, the moles of CaCl2 needed are half the moles of AgCl.

Step 5: Calculate the moles of CaCl2: $\text{{moles of CaCl2}} = \frac{1}{2} \times \text{{moles of AgCl}} = \frac{1}{2} \times 0.03 \text{{ mol}} = 0.015 \text{{ mol}}$

Final Answer: 0.015

Why other options are wrong:
- Option A (0.03): This is the number of moles of AgCl, not CaCl2. It fails to account for the 1:2 stoichiometric ratio.
- Option C (0.045): This is an incorrect calculation and doesn't follow from the stoichiometry of the reaction.
- Option D (0.06): This is also an incorrect calculation and doesn't follow from the stoichiometry of the reaction.

Key Concepts:
- Stoichiometry
- Mole concept
- Balanced chemical equations

Memory Aids:
- Remember to always balance the chemical equation before performing stoichiometric calculations.

Quick Tips:
- Always check the mole ratio between reactants and products in a balanced chemical equation.

48 Medium

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Application +4 | -1

The volume of air needed for complete combustion of 1 kg carbon at STP is:

View answer and explanation

Correct option: A

9333.33 litre

Option A (9333.33 litre) is correct because it accurately calculates the volume of air needed for complete combustion of 1 kg of carbon at STP, considering the oxygen content in air.

Solution approach

Correct Answer + Core Reason: 9333.33 litre. This is calculated by finding the volume of oxygen required to burn 1kg of carbon and then accounting for the fact that air is only ~21% oxygen.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the combustion of carbon.
$\mathrm{C + O_2 \rightarrow CO_2}$

Step 2: Determine the molar mass of carbon.\nThe molar mass of carbon (C) is 12 g/mol.

Step 3: Calculate the number of moles of carbon in 1 kg.
1 kg = 1000 g\nMoles of C = $\frac{1000 \text{ g}}{12 \text{ g/mol}} = 83.33 \text{ mol}$

Step 4: Determine the volume of oxygen required at STP.\nFrom the balanced equation, 1 mole of C requires 1 mole of $\mathrm{O_2}$ .\nSo, 83.33 moles of C require 83.33 moles of $\mathrm{O_2}$ .\nAt STP, 1 mole of any gas occupies 22.4 L.\nVolume of $\mathrm{O_2}$ = 83.33 mol $\times$ 22.4 L/mol = 1866.67 L

Step 5: Calculate the volume of air required.\nAir is approximately 21% oxygen by volume.\nVolume of air = $\frac{\text{Volume of } O_2}{0.21} = \frac{1866.67 \text{ L}}{0.2} = 9333.33 \text{ L}$

Why other options are wrong:
- Option B (933.33 litre): This is incorrect. It seems to be a miscalculation or not accounting for the percentage of oxygen in the air.
- Option C (93.33 litre): This is incorrect. It seems to be a miscalculation or not accounting for the percentage of oxygen in the air and the mass of carbon.
- Option D (1866.67 litre): This is the volume of oxygen required, but the question asks for the volume of air, which is larger because air is not pure oxygen.

Key Concepts:
- Stoichiometry
- Molar mass
- STP conditions
- Composition of air

Memory Aids:
- Remember that air is approximately 21% oxygen by volume.
- At STP, 1 mole of any gas occupies 22.4 L.

Quick Tips:
- Always write the balanced chemical equation first.
- Pay attention to the units and convert them if necessary.
- Remember to account for the composition of air when calculating the volume of air required.

49 Medium

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Application +4 | -1

6.90 grams of metal carbonate was completely decomposed by 50 ml of 2N HCl. The equivalent weight of metal is

View answer and explanation

Correct option: A

39

Option A (39) is correct because the equivalent weight of the metal is calculated by subtracting the equivalent weight of carbonate from the equivalent weight of the metal carbonate.

Solution approach

Correct Answer + Core Reason: 39. The equivalent weight of the metal is calculated using the neutralization reaction and equivalent weights.

Step-by-Step Solution:

Step 1: Calculate the number of equivalents of HCl used: \ $\frac{{NV}}{1000} = \frac{{2 \times 50}}{1000} = 0.1 \text{{ equivalents}}$ .

Step 2: The number of equivalents of metal carbonate is equal to the number of equivalents of HCl used: 0.1 equivalents.

Step 3: Calculate the equivalent weight of the metal carbonate: \ $\text{{Equivalent weight}} = \frac{{\text{{Weight of metal carbonate}}}}{{\text{{Number of equivalents}}}} = \frac{{6.90}}{0.1} = 69$ .

Step 4: Calculate the equivalent weight of the carbonate ion (CO3^2-): Equivalent weight = Molar mass / valency = 60 / 2 = 30.

Step 5: Calculate the equivalent weight of the metal: Equivalent weight of metal carbonate = Equivalent weight of metal + Equivalent weight of carbonate. Therefore, Equivalent weight of metal = 69 - 30 = 39.

Why other options are wrong:
- Option B (45): Incorrect calculation of the equivalent weight of the metal.
- Option C (72): Incorrect calculation of the equivalent weight of the metal.
- Option D (52): Incorrect calculation of the equivalent weight of the metal.

Key Concepts:
- Equivalent weight
- Neutralization reaction

Memory Aids:
- Number of equivalents = Normality * Volume (in liters)

Quick Tips:
- Remember to divide the molar mass by the valency to get the equivalent weight.
- Pay attention to the units.

50 Medium

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Analyzing +4 | -1

Choose the correct option based on following statements. Here ‘T’ stands for true and ‘F’ stands for false statement.
(i) Molecular mass of cane sugar

(C_{12}H_{22}O_{11})

is 182 amu.
(ii) 1 mole of cane sugar contains

6.022 \times 10^{23}

molecules of cane sugar.
(iii) 34.20 g of cane sugar contains

6.022 \times 10^{21}

molecules of cane sugar.

View answer and explanation

Correct option: C

FTF

Option C (FTF) is correct because the molecular mass of cane sugar is 342 amu, 1 mole contains 6.022 × 10²³ molecules, and 34.20 g contains 6.022 × 10²² molecules.

Solution approach

Correct Answer + Core Reason: FTF. The molecular mass of cane sugar is 342 amu, and 34.20 g contains 6.022 × 10²² molecules.

Step-by-Step Solution:

Step 1: Calculate the molecular mass of cane sugar ( $C_{12}H_{22}O_{11}$ ):
$12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342 \text{{ amu}}$ \nStatement (i) is False.

Step 2: State the number of molecules in 1 mole of cane sugar:
1 mole of any substance contains $6.022 \times 10^{23}$ molecules (Avogadro's number).\nStatement (ii) is True.

Step 3: Calculate the number of molecules in 34.20 g of cane sugar:\nMolar mass of cane sugar = 342 g/mol\nNumber of moles in 34.20 g = $\frac{34.20}{342} = 0.1 \text{{ mol}}$ \nNumber of molecules = $0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22} \text{{ molecules}}$ \nStatement (iii) is False.

Step 4: Determine the correct option: The correct option is FTF.

Why other options are wrong:
- Option A: TTF: Incorrect, statement (i) is false.
- Option B: TFT: Incorrect, statement (i) and (iii) are false.
- Option D: FTT: Incorrect, statement (iii) is false.

Key Concepts:
- Molecular mass
- Mole concept
- Avogadro's number

Memory Aids:
- Avogadro's number: $6.022 \times 10^{23}$

Quick Tips:
- Always calculate the molecular mass correctly before proceeding with mole calculations.

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Analyzing +4 | -1

The following sequence of reaction occurs in commercial production of aqueous nitric acid.

\begin{array}{cc} 4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) & \Delta \mathrm{H}=-904 \mathrm{~kJ}.........(i) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_2(\mathrm{~g}) & \Delta \mathrm{H}=-112 \mathrm{~kJ}.........(ii) \\ 3 \mathrm{NO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_3(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) & \Delta \mathrm{H}=-140 \mathrm{~kJ}.........(iii) \end{array}

Determine the total heat (in

\mathrm{kJ} / \mathrm{mole}

) liberated at constant pressure for the production of exactly 1 mole of aqueous nitric acid by this process.

View answer and explanation

Correct option: A

493

Option A (493) is correct because the total heat liberated is calculated by summing the enthalpy changes for the reactions required to produce 1 mole of aqueous nitric acid.

Solution approach

Correct Answer + Core Reason: 493. The heat liberated is calculated based on the stoichiometry of the reactions to produce 1 mole of $\mathrm{HNO}_3$ .

Step-by-Step Solution:

Step 1: Determine the stoichiometric relationships: To produce 2 moles of $\mathrm{HNO}_3$ , 3 moles of $\mathrm{NO}_2$ are required. To produce 3 moles of $\mathrm{NO}_2$ , 3/2 moles of $\mathrm{O}_2$ and 3/2 moles of $\mathrm{NO}$ are required. To produce 3/2 moles of $\mathrm{NO}$ , 3/2 moles of $\mathrm{NH}_3$ are required.

Step 2: Calculate the enthalpy change for each reaction per mole of $\mathrm{HNO}_3$ produced:
- Reaction (i): $\frac{3}{2} \times \frac{1}{4} \times (-904 \text{{ kJ}}) = -339 \text{{ kJ}}$
- Reaction (ii): $\frac{3}{2} \times \frac{1}{2} \times (-112 \text{{ kJ}}) = -84 \text{{ kJ}}$
- Reaction (iii): $\frac{1}{2} \times (-140 \text{{ kJ}}) = -70 \text{{ kJ}}$

Step 3: Sum the enthalpy changes: $-339 \text{{ kJ}} - 84 \text{{ kJ}} - 70 \text{{ kJ}} = -493 \text{{ kJ}}$

Step 4: The total heat liberated is the absolute value of the enthalpy change: $493 \text{{ kJ}}$

Why other options are wrong:
- Option B: Incorrect calculation of the heat liberated.
- Option C: Incorrect calculation of the heat liberated.
- Option D: Incorrect calculation of the heat liberated.

Key Concepts:
- Hess's Law
- Stoichiometry
- Enthalpy Change

Memory Aids:
- Remember to adjust the enthalpy change based on the stoichiometry of the reaction.

Quick Tips:
- Balance the chemical equations and carefully track the moles of each reactant and product.

52 Medium

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Application +4 | -1

The mass of 11.2 L of ammonia gas at STP is

View answer and explanation

Correct option: A

8.5 g

Option A (8.5 g) is correct because it is calculated by using the molar mass of ammonia and the volume at STP.

Solution approach

Correct Answer + Core Reason: 8.5 g. The mass of ammonia gas is calculated using the molar volume at STP.

Step-by-Step Solution:

Step 1: Recall the molar volume of a gas at STP: 22.4 L/mol

Step 2: Calculate the number of moles of ammonia ( $NH_3$ ) gas: Moles = Volume / Molar volume = 11.2 L / 22.4 L/mol = 0.5 mol

Step 3: Calculate the molar mass of ammonia ( $NH_3$ ): Molar mass = 14 (N) + 3(1) (H) = 17 g/mol

Step 4: Calculate the mass of ammonia gas: Mass = Moles x Molar mass = 0.5 mol x 17 g/mol = 8.5 g

Why other options are wrong:
- Option B (85 g): Incorrect. This is 10 times the correct mass.
- Option C (17 g): Incorrect. This is the mass of 1 mole of ammonia, not 0.5 moles.
- Option D (1.7 g): Incorrect. This is 1/10th of the molar mass.

Key Concepts:
- Molar volume at STP
- Mole concept
- Molar mass

Memory Aids:
- Remember the molar volume at STP: 22.4 L/mol
- Moles = Volume / Molar volume
- Mass = Moles x Molar mass

Quick Tips:
- Always use the molar volume at STP when calculating the number of moles of a gas at STP.
- Ensure you use the correct molar mass for the substance.

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Understanding +4 | -1

45.4 \mathrm{~L}

of dinitrogen reacted with

22.7 \mathrm{~L}

of dioxygen and

45.4 \mathrm{~L}

of nitrous oxide was formed. The reaction is given below.

2 \mathrm{~N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}(\mathrm{g})

Which law is being obeyed in this experiment?

View answer and explanation

Correct option: D

Gay Lussac's law of gaseous volumes

Option D (Gay Lussac's law of gaseous volumes) is correct because the volumes of the reacting gases and the product bear a simple whole-number ratio to each other, which is consistent with Gay-Lussac's Law.

Solution approach

Correct Answer + Core Reason: Gay Lussac's law of gaseous volumes. The volumes of the gases react in a simple whole number ratio.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2 \mathrm{N}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \longrightarrow 2 \mathrm{N}_2 \mathrm{O}(\mathrm{g})$

Step 2: Determine the volume ratio of reactants and products. The volume ratio is 2:1:2.

Step 3: State Gay-Lussac's law of gaseous volumes. Gay-Lussac's law states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all volumes are measured at the same temperature and pressure.

Step 4: Relate the experimental data to Gay-Lussac's law. The volumes of dinitrogen, dioxygen, and nitrous oxide are in the ratio 45.4 L : 22.7 L : 45.4 L, which simplifies to 2:1:2. This is a simple whole-number ratio.

Step 5: Conclude that the experiment obeys Gay-Lussac's law of gaseous volumes.

Why other options are wrong:
- Option A: Law of conservation of mass is always obeyed in chemical reactions, but it's not the primary law demonstrated by the volume ratios.
- Option B: Law of definite proportion applies to the mass ratios of elements in a compound, not the volume ratios of reacting gases.
- Option C: Law of multiple proportion applies when two elements form more than one compound, and the masses of one element that combine with a fixed mass of the other element are in a simple ratio. This is not directly applicable here.

Key Concepts:
- Gay-Lussac's law of gaseous volumes.
- Volume ratios in chemical reactions.

Memory Aids:
- Gay-Lussac's law deals with volume ratios of gases.

Quick Tips:
- Remember that Gay-Lussac's law applies to gases at the same temperature and pressure.

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Application +4 | -1

10 g of hydrogen and 64 of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:

View answer and explanation

Correct option: D

4 moles

Option D (4 moles) is correct because oxygen is the limiting reactant, and the stoichiometry of the reaction indicates that 1 mole of oxygen produces 2 moles of water.

Solution approach

Correct Answer + Core Reason: 4 moles. Oxygen is the limiting reactant, and 1 mole of oxygen produces 2 moles of water.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2H_2 + O_2 \rightarrow 2H_2O$

Step 2: Calculate the number of moles of hydrogen: $n_{H_2} = \frac{{10 \text{{ g}}}}{{2 \text{{ g/mol}}}} = 5 \text{{ moles}}$

Step 3: Calculate the number of moles of oxygen: $n_{O_2} = \frac{{64 \text{{ g}}}}{{32 \text{{ g/mol}}}} = 2 \text{{ moles}}$

Step 4: Determine the limiting reactant: From the balanced equation, 2 moles of $H_2$ react with 1 mole of $O_2$ . Therefore, 5 moles of $H_2$ would require 2.5 moles of $O_2$ . Since we only have 2 moles of $O_2$ , oxygen is the limiting reactant.

Step 5: Calculate the amount of water produced: From the balanced equation, 1 mole of $O_2$ produces 2 moles of $H_2O$ . Therefore, 2 moles of $O_2$ will produce $2 \times 2 = 4 \text{{ moles}}$ of $H_2O$ .

Why other options are wrong:
- Option A (1 mole): This is incorrect because it doesn't account for the stoichiometry of the reaction and the limiting reactant.
- Option B (2 moles): This is incorrect because it doesn't account for the stoichiometry of the reaction and the limiting reactant.
- Option C (3 moles): This is incorrect because it doesn't account for the stoichiometry of the reaction and the limiting reactant.

Key Concepts:
- Stoichiometry
- Limiting reactant
- Mole concept

Memory Aids:
- Remember to always balance the chemical equation before performing stoichiometric calculations.
- The limiting reactant is the reactant that is completely consumed in the reaction.

Quick Tips:
- Always identify the limiting reactant before calculating the amount of product formed.
- Use the mole ratio from the balanced equation to determine the amount of product formed.

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Analyzing +4 | -1

Assertion: 1 equivalent of

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

has 1 equivalent of

\mathrm{K}, \mathrm{Cr}

and

\mathrm{O}

each.
Reason: Equivalent and milliequivalent reacts in equal number to give same equivalent of product.

View answer and explanation

Correct option: C

Both Assertion and Reason are correct and Reason is correct explanation of Assertion

Option C is correct because both the assertion and reason are true, and the reason correctly explains why equivalents react in equal numbers.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct and Reason is correct explanation of Assertion. Equivalent and milliequivalent reacts in equal number to give same equivalent of product.

Step-by-Step Solution:

Step 1: Understand the concept of equivalents. An equivalent is the amount of a substance that will react with or supply one mole of hydrogen ions in an acid-base reaction or one mole of electrons in a redox reaction.

Step 2: Analyze the assertion. The assertion states that 1 equivalent of K2Cr2O7 has 1 equivalent of K, Cr, and O each. This is true because the number of equivalents is based on the reactive part of the molecule.

Step 3: Analyze the reason. The reason states that equivalent and milliequivalent react in equal number to give same equivalent of product. This is a fundamental principle of equivalent weights.

Step 4: Connect the assertion and reason. The reason explains why the assertion is true. The fact that equivalents react in equal numbers is the basis for the statement about K2Cr2O7.

Why other options are wrong:
- Option A: Assertion is correct but Reason is wrong: The reason is also correct.
- Option B: Assertion is wrong but Reason is correct: The assertion is also correct.
- Option D: Both Assertion and Reason are correct but Reason is not correct explanation of Assertion: The reason is the correct explanation of the assertion.

Key Concepts:
- Equivalent weight
- Stoichiometry

Memory Aids:
- Equivalents always react in equal numbers.

Quick Tips:
- Remember the definition of equivalent weight and how it relates to stoichiometry.

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Maximum number of atoms are present in

View answer and explanation

Correct option: D

1 \mathrm{~L} \mathrm{~H}_2 \mathrm{O}(\mathrm{l})

The correct answer is D (

1 \mathrm{~L} \mathrm{~H}_2 \mathrm{O}(\mathrm{l})

) because it contains the largest number of moles of water, and thus the largest number of atoms.

Solution approach

Correct Answer + Core Reason: $1 \mathrm{~L} \mathrm{~H}_2 \mathrm{O}(\mathrm{l})$ . It contains the largest number of atoms.

Step-by-Step Solution:

Step 1: Calculate the number of moles of atoms in each option.

Step 2: A) $1 \mathrm{~g}$ atom $\mathrm{Fe} = 1 \mathrm{~mol}$ atoms.

Step 3: B) $1 \mathrm{~g}$ molecule $\mathrm{H}_2 \mathrm{O} = 1 \mathrm{~mol}$ molecules $= 3 \mathrm{~mol}$ atoms.

Step 4: C) $16 \mathrm{~g} \mathrm{~CH}_4 = \frac{{16}}{{16}} = 1 \mathrm{~mol}$ molecules $= 5 \mathrm{~mol}$ atoms.

Step 5: D) $1 \mathrm{~L} \mathrm{~H}_2 \mathrm{O} = 1000 \mathrm{~mL} \mathrm{~H}_2 \mathrm{O} = \frac{{1000}}{{18}} = 55.5 \mathrm{~mol}$ molecules $= 55.5 \times 3 = 166.5 \mathrm{~mol}$ atoms.

Step 6: Compare the number of moles of atoms in each option.

Why other options are wrong:
- Option A: This is incorrect because it has the fewest number of atoms.
- Option B: This is incorrect because it has fewer atoms than option D.
- Option C: This is incorrect because it has fewer atoms than option D.

Key Concepts:
- Mole concept
- Avogadro's number

Memory Aids:
- 1 mole = $6.022 \times 10^{23}$ particles

Quick Tips:
- Convert mass to moles and then multiply by the number of atoms per molecule.

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Application +4 | -1

2 g of metal carbonate is neutralised completely by 100mL of 0.1 (N) HCl. The equivalent weight of metal carbonate is

View answer and explanation

Correct option: D

200

The correct answer is D (200) because the equivalent weight of the metal carbonate is calculated based on the neutralization reaction with HCl.

Solution approach

Correct Answer + Core Reason: 200. The equivalent weight is calculated based on the neutralization reaction.

Step-by-Step Solution:

Step 1: Calculate the number of gram equivalents of HCl: Number of gram equivalents of $\mathrm{HCl} = \frac{{100 \times 0.1}}{{1000}} = 0.01$

Step 2: Equate the number of gram equivalents of HCl and metal carbonate: Number of gram equivalents of HCl=Number of gram equivalents of metal carbonate = 0.01

Step 3: Calculate the mass of 1 g equivalent of the carbonate salt: Mass of 1 g equivalents of carbonate salt $= \frac{2}{{0.01}} = 200\\;g$

Step 4: State the equivalent mass of metal carbonate: Equivalent mass of metal carbonate $= 200$

Why other options are wrong:
- Option A (50): Incorrect calculation.
- Option B (100): Incorrect calculation.
- Option C (150): Incorrect calculation.

Key Concepts:
- Equivalent weight
- Neutralization reaction

Memory Aids:
- Remember the relationship between gram equivalents and normality.

Quick Tips:
- Always ensure the reaction is balanced before performing calculations.

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Out of

1.0 \mathrm{~g}

dioxygen,

1.0 \mathrm{~g}

atomic oxygen and

1.0 \mathrm{~g}

ozone, the maximum number of oxygen atoms are contained in

View answer and explanation

Correct option: D

All contain the same number of atoms

The correct answer is D (All contain the same number of atoms) because equal masses of the same element, regardless of their molecular form, contain the same number of atoms.

Solution approach

Correct Answer + Core Reason: All contain the same number of atoms. Since the mass of oxygen is the same in all cases, the number of oxygen atoms is the same.

Step-by-Step Solution:

Step 1: Calculate the number of moles of dioxygen ( $O_2$ ): $n_{O_2} = \frac{{1.0 \text{{ g}}}}{{32 \text{{ g/mol}}}} = \frac{{1}}{{32}} \text{{ mol}}$ .

Step 2: Calculate the number of oxygen atoms in $1.0 \text{{ g}}$ of $O_2$ : Number of atoms = $2 \times n_{O_2} \times N_A = 2 \times \frac{{1}}{{32}} \times N_A = \frac{{1}}{{16}} N_A$ .

Step 3: Calculate the number of moles of atomic oxygen ( $O$ ): $n_O = \frac{{1.0 \text{{ g}}}}{{16 \text{{ g/mol}}}} = \frac{{1}}{{16}} \text{{ mol}}$ .

Step 4: Calculate the number of oxygen atoms in $1.0 \text{{ g}}$ of $O$ : Number of atoms = $n_O \times N_A = \frac{{1}}{{16}} N_A$ .

Step 5: Calculate the number of moles of ozone ( $O_3$ ): $n_{O_3} = \frac{{1.0 \text{{ g}}}}{{48 \text{{ g/mol}}}} = \frac{{1}}{{48}} \text{{ mol}}$ .

Step 6: Calculate the number of oxygen atoms in $1.0 \text{{ g}}$ of $O_3$ : Number of atoms = $3 \times n_{O_3} \times N_A = 3 \times \frac{{1}}{{48}} \times N_A = \frac{{1}}{{16}} N_A$ .

Step 7: Compare the number of oxygen atoms in each case. They are all equal to $\frac{{1}}{{16}} N_A$ .

Why other options are wrong:
- Option A: Incorrect because all contain the same number of atoms.
- Option B: Incorrect because all contain the same number of atoms.
- Option C: Incorrect because all contain the same number of atoms.

Key Concepts:
- Mole concept
- Avogadro's number
- Number of atoms

Memory Aids:
- Remember that the number of atoms is proportional to the number of moles.

Quick Tips:
- Calculate the number of moles first.
- Multiply by the number of atoms per molecule and Avogadro's number.

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Arrange the following aq. dilute solution in the increasing order of vapour pressure
(a) 10 % (w / v) glucose

(b) 10 % (w / v) sucrose

(c) 10 % (w / v) urea

View answer and explanation

Correct option: D

(c)

<

(a)

<

(b)

Option A ((a) < (b) < (c)) is the correct order because vapor pressure is inversely proportional to the molecular weight of the solute.

Solution approach

Correct Answer + Core Reason: (a) < (b) < (c). Vapor pressure is inversely proportional to the molecular weight of the solute.

Step-by-Step Solution:

Step 1: Understand the relationship between vapor pressure and solute molecular weight.

Step 2: The formula for the relative lowering of vapor pressure is: $\frac{{P^\circ - P_s}}{{P^\circ}} = \frac{{n}}{{n + N}} \simeq \frac{{n}}{{N}} \simeq \frac{{w}}{{MW_{\text{{solute}}} \times N}}$

Step 3: Since the amount of solvent is the same for all solutions, the vapor pressure depends only on the molecular weight of the solute.

Step 4: Vapor pressure of solution is proportional to the molecular weight of the solute.

Step 5: The molecular weights are: Glucose (180), Sucrose (342), Urea (60).

Step 6: Therefore, the order of increasing vapor pressure is: Urea < Glucose < Sucrose, or (c) < (a) < (b).

Why other options are wrong:
- Option B: Incorrect order of vapor pressure based on molecular weight.
- Option C: Incorrect order of vapor pressure based on molecular weight.
- Option D: Incorrect order of vapor pressure based on molecular weight.

Key Concepts:
- Colligative properties
- Relative lowering of vapor pressure
- Molecular weight

Memory Aids:
- Vapor pressure decreases with increasing solute concentration and molecular weight.

Quick Tips:
- Remember that vapor pressure is inversely proportional to solute molecular weight.

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Application +4 | -1

Hydrazine reacts with

\mathrm{KIO}_3

in presence of

\mathrm{HCl}

as

\mathrm{N}_2 \mathrm{H}_4+\mathrm{IO}_3^{-}+2 \mathrm{H}^{+}+\mathrm{Cl}^{-} \longrightarrow \mathrm{ICl}+\mathrm{N}_2+3 \mathrm{H}_2 \mathrm{O}

The equivalent masses of

\mathrm{N}_2 \mathrm{H}_4

and

\mathrm{KIO}_3

respectively are:

View answer and explanation

Correct option: A

8 and 53.5

Option A (8 and 53.5) is correct because the equivalent mass of

N_2H_4

is calculated based on the change in oxidation number of nitrogen, and the equivalent mass of

KIO_3

is calculated based on the change in oxidation number of iodine.

Solution approach

Correct Answer + Core Reason: 8 and 53.5. Equivalent masses are calculated based on changes in oxidation numbers.

Step-by-Step Solution:

Step 1: Determine the change in oxidation number for $N_2H_4$ : The oxidation number of N in $N_2H_4$ is -2. In $N_2$ , it is 0. The change per nitrogen atom is 2. Since there are two nitrogen atoms, the total change is 4.

Step 2: Calculate the equivalent mass of $N_2H_4$ : Equivalent mass = $\frac{{\text{{Molar mass}}}}{{\text{{Total change in oxidation number}}}} = \frac{{32}}{{4}} = 8$

Step 3: Determine the change in oxidation number for $KIO_3$ : The oxidation number of I in $IO_3^-$ is +5. In ICl, it is +1. The change is 4.

Step 4: Calculate the equivalent mass of $KIO_3$ : Molar mass of $KIO_3$ = 39 + 127 + 3(16) = 214. Equivalent mass = $\frac{{214}}{{4}} = 53.5$

Step 5: Final Answer: 8 and 53.5

Why other options are wrong:
- Option B (16 and 53.5): Incorrect equivalent mass of $N_2H_4$ .
- Option C (8 and 35.6): Incorrect equivalent mass of $KIO_3$ .
- Option D (8 and 87): Incorrect equivalent mass of $KIO_3$ .

Key Concepts:
- Equivalent mass
- Oxidation number
- Redox reactions

Memory Aids:
- Remember that equivalent mass = $\frac{{\text{{Molar mass}}}}{{\text{{n-factor}}}}$

Quick Tips:
- Determine the change in oxidation number correctly.

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