An organic compound has an empirical formula ${\rm{(C}}{{\rm{H}}_2}{\rm{O)}}$ its vapour density is 45. The molecular formula of the compound is

${{\rm{C}}_3}{{\rm{H}}_6}{{\rm{O}}_3}$

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{O}}$

${{\rm{C}}_2}{{\rm{H}}_2}{\rm{O}}$

Which is correct for ${\rm{N}}{{\rm{a}}_2}{\rm{HP}}{{\rm{O}}_3}$?

${\rm{Eq}}{\rm{. wt}}{\rm{.}} = M/2$

SOME BASIC CONCEPTS OF CHEMISTRY MCQs for JEE CHEMISTRY — Medium, Page 6

151 Medium

151 of 272

Application +4 | -1

At constant temperature for 1 mole of an ideal gas, if its pressure is decreased by

20 \%

then its volume will be increased by

View answer and explanation

Correct option: C

25 \%

Option C (

25 \%

) is correct because when the pressure is decreased by 20%, the volume increases by 25% according to Boyle's Law.

Solution approach

Correct Answer + Core Reason: $25 \%$ . The volume increases by 25% when pressure decreases by 20% according to Boyle's Law.

Step-by-Step Solution:

Step 1: Apply Boyle's Law: $P_1V_1 = P_2V_2$ (at constant temperature and number of moles).

Step 2: Let the initial pressure be $P_1 = P$ and the initial volume be $V_1 = V$ .

Step 3: The pressure is decreased by 20%, so the new pressure is $P_2 = P - 0.20P = 0.80P$ .

Step 4: Substitute the values into Boyle's Law: $PV = (0.80P)V_2$

Step 5: Solve for $V_2$ : $V_2 = \frac{PV}{0.80P} = \frac{V}{0.80} = 1.25V$

Step 6: Calculate the increase in volume: Increase = $V_2 - V = 1.25V - V = 0.25V$

Step 7: Calculate the percentage increase in volume: Percentage increase = $\frac{0.25V}{V} \times 100 = 25 \%$

Why other options are wrong:
- Option A (20 %): This is incorrect because it assumes a direct linear relationship between pressure and volume, which is not the case according to Boyle's Law.
- Option B (10 %): This is incorrect and doesn't follow from Boyle's Law.
- Option D (50 %): This is incorrect and doesn't follow from Boyle's Law.

Key Concepts:
- Boyle's Law
- Ideal gas law

Memory Aids:
- Boyle's Law: $P_1V_1 = P_2V_2$

Quick Tips:
- Remember that Boyle's Law applies at constant temperature and number of moles.

152 Medium

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Application +4 | -1

Which one of the following sets of compounds correctly illustrate the law of reciprocal proportions?

View answer and explanation

Correct option: A

P_2O_3, PH_3, H_2O

Option A (

P_2O_3, PH_3, H_2O

) is correct because the ratios of elements combining with a fixed mass of a third element are related to the ratio in which they combine with each other.

Solution approach

Correct Answer + Core Reason: $P_2O_3, PH_3, H_2O$ . These compounds illustrate the law of reciprocal proportions.

Step-by-Step Solution:

Step 1: State the Law of Reciprocal Proportions: When two different elements combine separately with the same weight of a third element, the ratio of their masses in the first two compounds is the same or a simple multiple of the ratio of their masses in which they combine with each other.

Step 2: Analyze $P_2O_3$ and $PH_3$ : In $P_2O_3$ , 2 moles of P combine with 3 moles of O. In $PH_3$ , 1 mole of P combines with 3 moles of H. Let's consider the ratio of H and O combining with a fixed amount of P.

Step 3: Analyze $H_2O$ : In $H_2O$ , 2 moles of H combine with 1 mole of O. The ratio of H to O is 2:1.

Step 4: Compare the ratios: The ratio of H:O combining with P is 3:1.5 or 2:1. This is the same ratio as in $H_2O$ .

Step 5: Conclusion: The compounds $P_2O_3, PH_3,$ and $H_2O$ illustrate the law of reciprocal proportions.

Why other options are wrong:
- Option B: $P_2O_5, PH_3, H_2O$ : The ratio of oxygen and hydrogen combining with phosphorus in $P_2O_5$ and $PH_3$ is not a simple multiple of the ratio in which they combine in $H_2O$ .
- Option C: $N_2O_5, NH_3, H_2O$ : The ratio of oxygen and hydrogen combining with nitrogen in $N_2O_5$ and $NH_3$ is not a simple multiple of the ratio in which they combine in $H_2O$ .
- Option D: $N_2O, NH_3, H_2O$ : The ratio of oxygen and hydrogen combining with nitrogen in $N_2O$ and $NH_3$ is not a simple multiple of the ratio in which they combine in $H_2O$ .

Key Concepts:
- Law of Reciprocal Proportions
- Chemical formulas and mole ratios

Memory Aids:
- Reciprocal Proportions: Remember that the ratios must be the same or a simple multiple.

Quick Tips:
- To check for reciprocal proportions, find a common element and compare the ratios of the other elements combining with it.

153 Medium

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Application +4 | -1

Oxidation number of carbon in carbon suboxide is :

View answer and explanation

Correct option: B

+ \frac{4}{3}

Option B (+4/3) is correct because it is calculated by setting up an equation where the sum of oxidation states equals zero.

Solution approach

Correct Answer + Core Reason: $+ \frac{4}{3}$ . The oxidation number of carbon is calculated by setting the sum of oxidation states to zero.

Step-by-Step Solution:

Step 1: Identify the compound: C₃O₂ (carbon suboxide)

Step 2: Assign oxidation numbers:
- Oxygen (O) usually has an oxidation number of -2.

Step 3: Let the oxidation number of carbon be 'x'. Set up the equation to solve for x:
3(x) + 2(-2) = 0

Step 4: Solve for x:
3x - 4 = 0
3x = 4
$x = +\frac{4}{3}$

Step 5: Final Answer: +4/3

Why other options are wrong:
- Option A (+2/3): This is incorrect because it involves an error in the calculation, possibly by not correctly accounting for the number of carbon atoms or by misinterpreting the formula.
- Option C (+4): This is incorrect because it does not correctly account for the number of carbon and oxygen atoms in the molecule and their respective oxidation states.
- Option D (-4/3): This is incorrect because the oxidation state of carbon should be positive in this compound, as oxygen is more electronegative.

Key Concepts:
- Oxidation Number
- Rules for Assigning Oxidation Numbers
- Electronegativity

Memory Aids:
- The sum of oxidation numbers in a neutral compound is zero.
- Oxygen usually has an oxidation number of -2.

Quick Tips:
- Remember to account for the number of atoms of each element when setting up the equation to solve for oxidation numbers.
- When assigning oxidation numbers, start with the more electronegative element.

154 Medium

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Application +4 | -1

1.5276 \mathrm{~g}

of

\mathrm{CdCl}_2

was found to contain

0.9367 \mathrm{~g}

of

\mathrm{Cd}

. The atomic mass of

\mathrm{Cd}

is (to the nearest integer):

View answer and explanation

Correct option: A

113 g

Option A (113 g) is correct because the atomic mass of Cd is calculated using the equivalent mass concept and the given masses of Cd and Cl in CdCl2.

Solution approach

Correct Answer + Core Reason: 113 g. The atomic mass of Cd is calculated using the equivalent mass concept.

Step-by-Step Solution:

Step 1: Calculate the mass of Cl in CdCl2: Mass of Cl = Mass of CdCl2 - Mass of Cd = 1.5276 g - 0.9367 g = 0.5909 g

Step 2: Apply the concept of equivalent mass: $\frac{\text{Mass of Cd}}{\text{Equivalent mass of Cd}} = \frac{\text{Mass of Cl}}{\text{Equivalent mass of Cl}}$

Step 3: Equivalent mass of Cl = Atomic mass of Cl = 35.5 g/mol

Step 4: Substitute the values: $\frac{0.9367}{\text{Equivalent mass of Cd}} = \frac{0.5909}{35.5}$

Step 5: Solve for the equivalent mass of Cd: Equivalent mass of Cd = $\frac{0.9367 \times 35.5}{0.5909} = 56.275$ g/equivalent

Step 6: Since Cd has a valency of 2 in CdCl2, Atomic mass of Cd = Equivalent mass of Cd * Valency = 56.275 * 2 = 112.55 g/mol

Step 7: Round to the nearest integer: Atomic mass of Cd ≈ 113 g/mol

Step 8: Final Answer: 113 g

Why other options are wrong:
- Option B: 111 g: Incorrect calculation or rounding error.
- Option C: 118 g: Incorrect calculation or incorrect application of the equivalent mass concept.
- Option D: 120 g: Incorrect calculation or incorrect application of the equivalent mass concept.

Key Concepts:
- Equivalent Mass
- Atomic Mass
- Stoichiometry

Memory Aids:
- Remember that Equivalent mass = Atomic mass / Valency

Quick Tips:
- Always identify the valency of the element before calculating the atomic mass.

155 Medium

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Application +4 | -1

In an experiment 50 mL of 0.1

M

solution of a salt reacted with 25 mL of 0.1

M

solution of sodium sulphite. The half equation for the oxidation of

\mathrm{SO}_3^{2 - }\left( {aq} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right) \to \mathrm{SO}_4^{2 - }\left( {aq} \right) + 2{\mathrm{H}^ + }\left( {aq} \right) + 2{e^ - }

If the oxidation number of metal in the salt was 3, what would be the new oxidation number of metal?

View answer and explanation

Correct option: C

2

Option C (2) is correct because the change in oxidation number of the metal is 1, resulting in a new oxidation number of 2.

Solution approach

Correct Answer + Core Reason: 2. The new oxidation number is calculated based on the stoichiometry of the reaction and the change in oxidation state.

Step-by-Step Solution:

Step 1: Determine the meq of Na2SO3: $Meq.\mathrm{ }of\mathrm{ }N{a_2}S{O_3} = 25 \times 0.1 \times 2 = 5$

Step 2: Determine the meq of the salt: $Meq.\mathrm{ }of\mathrm{ }salt = 50 \times 0.1 \times n = 5n$

Step 3: Equate the meq of salt and Na2SO3: $5n = 5$ , so $n = 1$

Step 4: Determine the new oxidation number: Since the oxidation number of the metal was 3 and the change is 1, the new oxidation number is $3 - 1 = 2$

Why other options are wrong:
- Option A: Zero is incorrect because the metal gains one electron, reducing its oxidation state from +3 to +2, not to zero.
- Option B: 1 is incorrect because the metal gains one electron, reducing its oxidation state from +3 to +2, not to +1.
- Option D: 4 is incorrect because the metal gains one electron, reducing its oxidation state from +3 to +2, not increasing it to +4.

Key Concepts:
- Redox reactions
- Oxidation number
- Stoichiometry

Memory Aids:
- Oxidation is Loss, Reduction is Gain (OIL RIG)

Quick Tips:
- Balance the equation to determine the number of electrons transferred.
- Remember that oxidation number decreases during reduction and increases during oxidation.

156 Medium

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Application +4 | -1

An aqueous solution of oxalic acid dihydrate contains its

6.3 \mathrm{~g}

in

250 \mathrm{~mL}

. The volume of

0.1 \mathrm{~N} \mathrm{~NaOH}

required to completely neutralize

10 \mathrm{~mL}

of this solution

View answer and explanation

Correct option: D

40 \mathrm{~mL}

Option D (40 mL) is correct because it is calculated using the normality equation N₁V₁ = N₂V₂ to determine the volume of NaOH required for complete neutralization.

Solution approach

Correct Answer + Core Reason: 40 mL. The volume of NaOH is calculated using the normality equation for neutralization.

Step-by-Step Solution:

Step 1: Calculate the molar mass of oxalic acid dihydrate (H₂C₂O₄·2H₂O).\nMolar mass = 2(1) + 2(12) + 4(16) + 2(2(1) + 16) = 2 + 24 + 64 + 36 = 126 g/mol

Step 2: Determine the number of moles of oxalic acid in 250 mL of solution.\nMoles of oxalic acid = $\frac{\text{mass}}{\text{molar mass}} = \frac{6.3}{126} = 0.05$ mol

Step 3: Calculate the molarity of the oxalic acid solution.\nMolarity = $\frac{\text{moles}}{\text{volume in L}} = \frac{0.05}{0.250} = 0.2$ M

Step 4: Determine the normality of the oxalic acid solution. Oxalic acid is a diprotic acid, so its normality is twice its molarity.\nNormality = 2 × Molarity = 2 × 0.2 = 0.4 N

Step 5: Use the neutralization equation to find the volume of NaOH required to neutralize 10 mL of the oxalic acid solution.
$N_1V_1 = N_2V_2$ , where $N_1$ is the normality of oxalic acid, $V_1$ is the volume of oxalic acid, $N_2$ is the normality of NaOH, and $V_2$ is the volume of NaOH.
$(0.4 N)(10 mL) = (0.1 N)(V_2)$

Step 6: Solve for $V_2$ .
$V_2 = \frac{(0.4)(10)}{0.1} = 40 mL$

Final Answer: 40 mL

Why other options are wrong:
- Option A (4 mL): This is likely a calculation error, possibly not accounting for the diprotic nature of oxalic acid.
- Option B (20 mL): This is a calculation error, likely using molarity instead of normality in the neutralization equation.
- Option C (2 mL): This is a significant calculation error, possibly not using the correct molarity or normality.

Key Concepts:
- Normality
- Molarity
- Neutralization reaction
- Stoichiometry

Memory Aids:
- Normality = n × Molarity, where n is the number of equivalents per mole.
- $N_1V_1 = N_2V_2$ for neutralization reactions.

Quick Tips:
- Remember to account for the number of acidic or basic protons when converting between molarity and normality.
- Ensure you use the correct units (mL or L) for volume.

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Application +4 | -1

Arsenic forms two oxides, one of which contains 65.2% and the other 75.5% of the element. Hence, equivalent masses of arsenic are in the ratio

View answer and explanation

Correct option: B

3\mathrm{ }:\mathrm{ }5

Option B (

3\mathrm{ }:\mathrm{ }5

) is correct because the equivalent mass is calculated by dividing the mass of the element by the mass of oxygen and multiplying by 8. The ratio of the equivalent masses is then simplified.

Solution approach

Correct Answer + Core Reason: $3\mathrm{ }:\mathrm{ }5$ . The ratio of equivalent masses is found by calculating the equivalent mass of arsenic in each oxide and then finding the ratio.

Step-by-Step Solution:

Step 1: Define equivalent mass:\nEquivalent mass = (Mass of element / Mass of oxygen) * 8

Step 2: Calculate the equivalent mass of arsenic in the first oxide:\nMass of arsenic = 65.2%\nMass of oxygen = 100% - 65.2% = 34.8%\nEquivalent mass of arsenic = (65.2 / 34.8) * 8 = 14.99

Step 3: Calculate the equivalent mass of arsenic in the second oxide:\nMass of arsenic = 75.7%\nMass of oxygen = 100% - 75.7% = 24.3%\nEquivalent mass of arsenic = (75.7 / 24.3) * 8 = 24.92

Step 4: Calculate the ratio of the equivalent masses:\nRatio = 14.99 : 24.92 ≈ 15 : 25 = 3 : 5

Why other options are wrong:
- **Option A ( $1\mathrm{ }:\mathrm{ }2$ ): This ratio is incorrect based on the given percentages of arsenic and oxygen in the two oxides.
- Option C ( $13\mathrm{ }:\mathrm{ }15$ ): This ratio is incorrect based on the given percentages of arsenic and oxygen in the two oxides.
- Option D ( $2\mathrm{ }:\mathrm{ }1$ ): This ratio is incorrect based on the given percentages of arsenic and oxygen in the two oxides.

Key Concepts:
- Equivalent mass
- Percentage composition

Memory Aids:**
- Equivalent mass = (Mass of element / Mass of oxygen) * 8

Quick Tips:
- Ensure the percentages add up to 100% before calculating the mass of oxygen.
- Simplify the ratio to its lowest terms.

158 Medium

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Application +4 | -1

20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?
(Atomic weight:

Mg = 24

)

View answer and explanation

Correct option: B

84

Option B (84) is correct because the percentage purity is calculated by comparing the actual yield of MgO to the theoretical yield based on the decomposition of MgCO3.

Solution approach

Correct Answer + Core Reason: 84%. The percentage purity of magnesium carbonate is calculated based on the mass of magnesium oxide produced.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the decomposition of magnesium carbonate: $MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$ .

Step 2: Calculate the molar mass of $MgCO_3$ : $24 + 12 + (3 \times 16) = 84 \text{{ g/mol}}$ .

Step 3: Calculate the moles of $MgCO_3$ in the sample: $\frac{20 \text{{ g}}}{84 \text{{ g/mol}}} = 0.238 \text{{ mol}}$ .

Step 4: Calculate the molar mass of $MgO$ : $24 + 16 = 40 \text{{ g/mol}}$ .

Step 5: Calculate the theoretical mass of $MgO$ produced from 0.238 mol of $MgCO_3$ : $0.238 \text{{ mol}} \times 40 \text{{ g/mol}} = 9.52 \text{{ g}}$ .

Step 6: Calculate the percentage purity of $MgCO_3$ in the sample: $\frac{\text{{Actual yield of MgO}}}{\text{{Theoretical yield of MgO}}} \times 100 = \frac{8 \text{{ g}}}{9.52 \text{{ g}}} \times 100 = 84\%$ .

Step 7: Final Answer: 84%

Why other options are wrong:
- Option A (60): This is incorrect because the calculation is wrong. The correct calculation yields 84%.
- Option C (75): This is incorrect because the calculation is wrong. The correct calculation yields 84%.
- Option D (96): This is incorrect because the calculation is wrong. The correct calculation yields 84%.

Key Concepts:
- Stoichiometry
- Percentage purity
- Decomposition reactions

Memory Aids:
- Remember the formula for percentage purity: $\frac{\text{{Actual yield}}}{\text{{Theoretical yield}}} \times 100$

Quick Tips:
- Always write the balanced chemical equation before performing stoichiometric calculations.
- Make sure to use the correct molar masses for all compounds.

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Application +4 | -1

Dissolving 120 g of urea (mol.wt.60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is:

View answer and explanation

Correct option: C

2.05 M

Option C (2.05 M) is correct because it is calculated using the formula for molarity, considering the mass of solute, volume of solution, and density.

Solution approach

Correct Answer + Core Reason: 2.05 M. The molarity is calculated by dividing the moles of solute by the volume of the solution in liters.

Step-by-Step Solution:

Step 1: Calculate the moles of urea.\nMoles of urea = $\frac{{\text{{mass of urea}}}}{{\text{{molar mass of urea}}}} = \frac{{120 \text{{ g}}}}{{60 \text{{ g/mol}}}} = 2 \text{{ mol}}$

Step 2: Calculate the mass of the solution.\nMass of solution = Mass of urea + Mass of water = 120 g + 1000 g = 1120 g

Step 3: Calculate the volume of the solution.\nVolume of solution = $\frac{{\text{{mass of solution}}}}{{\text{{density of solution}}}} = \frac{{1120 \text{{ g}}}}{{1.15 \text{{ g/mL}}}} = 973.91 \text{{ mL}}$

Step 4: Convert the volume of the solution to liters.\nVolume of solution = $\frac{{973.91 \text{{ mL}}}}{{1000 \text{{ mL/L}}}} = 0.97391 \text{{ L}}$

Step 5: Calculate the molarity of the solution.\nMolarity = $\frac{{\text{{moles of solute}}}}{{\text{{volume of solution in liters}}}} = \frac{{2 \text{{ mol}}}}{{0.97391 \text{{ L}}}} = 2.05 \text{{ M}}$

Why other options are wrong:
- Option A (1.78 M): This is incorrect because it likely uses the mass of water instead of the volume of the solution to calculate molarity.
- Option B (2.00 M): This is incorrect due to rounding errors or incorrect calculation of the solution volume.
- Option D (2.22 M): This is incorrect and likely results from an error in calculating or using the solution volume.

Key Concepts:
- Molarity
- Density
- Mole concept

Memory Aids:\nMolarity = Moles of solute / Liters of solution

Quick Tips:
- Always convert the volume to liters before calculating molarity.
- Make sure to use the volume of the *solution*, not just the volume of the solvent.

160 Medium

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Application +4 | -1

Mass of

{10^{21}}

molecules of Cu (at. Wt = 63.5)

View answer and explanation

Correct option: A

0.106

Option A (0.106) is correct because it is the mass calculated by finding the mass of one molecule and multiplying by the number of molecules.

Solution approach

Correct Answer + Core Reason: 0.106. The mass is calculated using Avogadro's number and the atomic weight.

Step-by-Step Solution:

Step 1: Identify the given information.\nNumber of molecules = ${10^{21}}$ \nAtomic weight of Cu = 63.5 g/mol\nAvogadro's number = $6.022 \times {10^{23}}$ molecules/mol

Step 2: Calculate the mass of one mole of Cu.\nMass of one mole of Cu = 63.5 g

Step 3: Calculate the mass of one molecule of Cu.\nMass of one molecule of Cu = $\frac{{63.5}}{{6.022 \times {{10}^{23}}}} g$

Step 4: Calculate the mass of ${10^{21}}$ molecules of Cu.\nMass of ${10^{21}}$ molecules of Cu = $\frac{{63.5}}{{6.022 \times {{10}^{23}}}} \times {10^{21}} g$

Step 5: Simplify the expression.\nMass of ${10^{21}}$ molecules of Cu = $\frac{{63.5}}{{602.2}} g \approx 0.106 g$

Why other options are wrong:
- Option B (10.6 g): This is incorrect because it likely involves multiplying 63.5 by ${10^{21}}$ and dividing by ${10^{23}}$ incorrectly.
- Option C (63.5 g): This is incorrect as it represents the mass of one mole of Cu, not ${10^{21}}$ molecules.
- Option D (106 g): This is incorrect and suggests a significant error in the calculation.

Key Concepts:
- Moles
- Avogadro's number
- Atomic weight

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

Quick Tips:
- Remember Avogadro's number.
- Ensure you are using the correct units.

161 Medium

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Application +4 | -1

Mole fraction of solute in aqueous solution of

30 \% \,{NaOH}

.

View answer and explanation

Correct option: A

0.16

Option A (0.16) is correct because it is calculated by finding the moles of NaOH and water, then finding the mole fraction of NaOH.

Solution approach

Correct Answer + Core Reason: 0.16. The mole fraction is calculated by dividing the moles of NaOH by the total moles of NaOH and water.

Step-by-Step Solution:

Step 1: Understand the meaning of 30% NaOH aqueous solution. It means 30 g of NaOH is present in 100 g of solution.

Step 2: Calculate the mass of water in the solution. Mass of water = 100 g (solution) - 30 g (NaOH) = 70 g.

Step 3: Calculate the number of moles of NaOH. Moles of NaOH = $\frac{{\text{{mass of NaOH}}}}{{\text{{molar mass of NaOH}}}} = \frac{{30 \text{{ g}}}}{{40 \text{{ g/mol}}}} = 0.75 \text{{ mol}}$

Step 4: Calculate the number of moles of water. Moles of water = $\frac{{\text{{mass of water}}}}{{\text{{molar mass of water}}}} = \frac{{70 \text{{ g}}}}{{18 \text{{ g/mol}}}} = 3.89 \text{{ mol}}$

Step 5: Calculate the mole fraction of NaOH. Mole fraction of NaOH = $\frac{{\text{{moles of NaOH}}}}{{\text{{moles of NaOH}} + \text{{moles of water}}}} = \frac{{0.75}}{{0.75 + 3.89}} = \frac{{0.75}}{{4.64}} = 0.16$

Why other options are wrong:
- Option B: Incorrect calculation. This value is likely obtained by an incorrect application of the mole fraction formula or using incorrect masses.
- Option C: Incorrect calculation. This value is likely obtained by an incorrect application of the mole fraction formula or using incorrect masses.
- Option D: Incorrect calculation. This value is likely obtained by an incorrect application of the mole fraction formula or using incorrect masses.

Key Concepts:
- Mole fraction
- Molar mass
- Percentage composition

Memory Aids:
- Mole fraction = (moles of component) / (total moles in solution)

Quick Tips:
- Always convert mass to moles before calculating mole fraction.
- Make sure to use the correct molar masses.

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Analyzing +4 | -1

A compound has molar mass of

147 \mathrm{~g}

. It contains

49 \%\mathrm{C}

and

2.72 \% \mathrm{H}

. On Carius estimation

2.561 \mathrm{~mg}

of the compound gave

5 \mathrm{~mg}

of

\mathrm{AgCl}

. Find the empirical and molecular formula.

View answer and explanation

Correct option: A

\mathrm{C}_3 \mathrm{H}_2 \mathrm{Cl}, \mathrm{C}_6 \mathrm{H}_4 \mathrm{Cl}_2

Option A (

\mathrm{C}_3 \mathrm{H}_2 \mathrm{Cl}, \mathrm{C}_6 \mathrm{H}_4 \mathrm{Cl}_2

) is correct because the empirical formula is calculated from the percentage composition and the molecular formula is a multiple of the empirical formula.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}_3 \mathrm{H}_2 \mathrm{Cl}, \mathrm{C}_6 \mathrm{H}_4 \mathrm{Cl}_2$

Step-by-Step Solution:

Step 1: Calculate the percentage of Cl in the compound using Carius estimation:
$\text{{%Cl}} = \frac{{35.5}}{{143.5}} \times \frac{{\text{{mass of AgCl}}}}{{\text{{mass of compound}}}} \times 100$
$\text{{%Cl}} = \frac{{35.5}}{{143.5}} \times \frac{{5 \text{{ mg}}}}{{2.581 \text{{ mg}}}}\times 100 = 47.9 \%$ (approximately)

Step 2: Determine the atomic ratio of each element:\nC: $\frac{{49}}{{12}} = 4.08 \approx 4.1$ \nH: $\frac{{2.72}}{{1}} = 2.72$ \nCl: $\frac{{47.9}}{{35.5}} = 1.35 \approx 1.34$

Step 3: Find the simplest whole number ratio by dividing each atomic ratio by the smallest value (1.34):\nC: $\frac{{4.1}}{{1.34}} \approx 3$ \nH: $\frac{{2.72}}{{1.34}} \approx 2$ \nCl: $\frac{{1.34}}{{1.34}} = 1$

Step 4: The empirical formula is $\mathrm{C}_3 \mathrm{H}_2 \mathrm{Cl}$ .

Step 5: Calculate the empirical formula mass:
$3(12) + 2(1) + 1(35.5) = 36 + 2 + 35.5 = 73.5 \text{{ g/mol}}$

Step 6: Determine the value of 'n' by dividing the molar mass by the empirical formula mass:\nn = $\frac{{147 \text{{ g/mol}}}}{{73.5 \text{{ g/mol}}}} = 2$

Step 7: Calculate the molecular formula by multiplying the empirical formula by 'n':
$(\mathrm{C}_3 \mathrm{H}_2 \mathrm{Cl})_2 = \mathrm{C}_6 \mathrm{H}_4 \mathrm{Cl}_2$

Why other options are wrong:
- Option B: Incorrect because the empirical and molecular formulas are not calculated correctly.
- Option C: Incorrect because the empirical and molecular formulas are not calculated correctly.
- Option D: Incorrect because the empirical and molecular formulas are not calculated correctly.

Key Concepts:
- Empirical Formula
- Molecular Formula
- Percentage Composition
- Carius Method

Memory Aids:
- Remember the steps for determining empirical and molecular formulas.

Quick Tips:
- Always check if the given percentages add up to 100%. If not, consider oxygen.
- Use the Carius method formula to calculate the percentage of halogen.

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Application +4 | -1

The number of

\mathrm{F}{\mathrm{e}^{2 + }}

ion oxidised by one mole of

\mathrm{MnO}_4^ -

ions is :

View answer and explanation

Correct option: C

5

Option C (5) is correct because one mole of

\mathrm{MnO}_4^ -

gains 5 electrons, and each

\mathrm{F}{\mathrm{e}^{2 + }}

loses one electron during oxidation.

Solution approach

Correct Answer + Core Reason: 5. The stoichiometry of the redox reaction dictates that 5 moles of $\mathrm{F}{\mathrm{e}^{2 + }}$ are oxidized per mole of $\mathrm{MnO}_4^ -$ .

Step-by-Step Solution:

Step 1: Write the half-reactions for the oxidation and reduction processes.

Step 2: Reduction half-reaction: $\mathrm{MnO}_4^ - + 8{\mathrm{H}^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$

Step 3: Oxidation half-reaction: $\mathrm{F}{\mathrm{e}^{2 + }} \to \mathrm{F}{\mathrm{e}^{3 + }} + {e^ - }$

Step 4: Balance the electron transfer. For every one $\mathrm{MnO}_4^ -$ that is reduced, 5 $\mathrm{F}{\mathrm{e}^{2 + }}$ ions are oxidized.

Step 5: Therefore, the number of $\mathrm{F}{\mathrm{e}^{2 + }}$ ions oxidized by one mole of $\mathrm{MnO}_4^ -$ ions is 5.

Why other options are wrong:
- **Option A ( $\frac{1}{5}$ ):** This is the reciprocal of the correct ratio. It represents the moles of $\mathrm{MnO}_4^ -$ needed to oxidize one mole of $\mathrm{F}{\mathrm{e}^{2 + }}$ .
- **Option B ( $\frac{2}{3}$ ): This value is incorrect and doesn't relate to the electron transfer stoichiometry.
- Option D ( $\frac{3}{2}$ ): This value is also incorrect and doesn't relate to the electron transfer stoichiometry.

Key Concepts:
- Redox reactions
- Balancing redox reactions
- Stoichiometry

Memory Aids:
- OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons)

Quick Tips:**
- Always balance the redox reactions to determine the correct stoichiometry.

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Application +4 | -1

An organic compound has an empirical formula

\mathrm{(C}{\mathrm{H}_2}\mathrm{O)}

its vapour density is 45. The molecular formula of the compound is

View answer and explanation

Correct option: D

{\mathrm{C}_3}{\mathrm{H}_6}{\mathrm{O}_3}

Option D (

{\mathrm{C}_3}{\mathrm{H}_6}{\mathrm{O}_3}

) is correct because the molecular formula is determined by multiplying the empirical formula by a factor derived from the vapor density.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_3}{\mathrm{H}_6}{\mathrm{O}_3}$ . The molecular formula is found by comparing the empirical formula weight to the molecular weight calculated from vapor density.

Step-by-Step Solution:

Step 1: Calculate the molecular weight using the vapor density: Molecular weight = $2 \times$ vapor density = $2 \times 45 = 90$

Step 2: Calculate the empirical formula weight: Empirical formula weight of $\mathrm{C}{\mathrm{H}_2}\mathrm{O} = 12 + 2 + 16 = 30$

Step 3: Calculate the factor 'n': $n = \frac{{\text{{Molecular weight}}}}{{\text{{Empirical formula weight}}}} = \frac{{90}}{{30}} = 3$

Step 4: Determine the molecular formula: Molecular formula = $(\mathrm{C}{\mathrm{H}_2}\mathrm{O})_3 = {\mathrm{C}_3}{\mathrm{H}_6}{\mathrm{O}_3}$

Why other options are wrong:
- **Option A ( $\mathrm{C}{\mathrm{H}_2}\mathrm{O}$ ): This is the empirical formula, not the molecular formula.
- Option B ( ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{O}$ ): Incorrectly calculated molecular formula.
- Option C ( ${\mathrm{C}_2}{\mathrm{H}_2}\mathrm{O}$ ): Incorrectly calculated molecular formula.

Key Concepts:
- Empirical formula
- Molecular formula
- Vapor density

Memory Aids:**
- Molecular weight = $2 \times$ Vapor Density
- Molecular formula = (Empirical formula) $_n$

Quick Tips:
- Always calculate the empirical formula weight and compare it to the molecular weight to find the correct 'n' value.

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Application +4 | -1

12 \mathrm{gm}

of

\mathrm{Mg}

(At mass

=24

) will react with an acid to give

View answer and explanation

Correct option: C

half moles of

\mathrm{H}_2

Option C (half moles of

\mathrm{H}_2

) is correct because 12 gm of Mg (half a mole) will react with an acid to produce half a mole of

\mathrm{H}_2

.

Solution approach

Correct Answer + Core Reason: half moles of $\mathrm{H}_2$ . The reaction of Mg with acid produces $\mathrm{H}_2$ gas, and the stoichiometry dictates the amount produced.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the reaction of magnesium with an acid:
$\mathrm{Mg} + 2\mathrm{H}^+ \longrightarrow \mathrm{Mg}^{2+} + \mathrm{H}_2$

Step 2: Determine the number of moles of Mg used in the reaction:\nMoles of Mg = $\frac{\text{mass of Mg}}{\text{atomic mass of Mg}} = \frac{12 \text{ gm}}{24 \text{ gm/mol}} = 0.5 \text{ mol}$

Step 3: Use the stoichiometry of the balanced equation to determine the number of moles of $\mathrm{H}_2$ produced.\nFrom the balanced equation, 1 mole of Mg produces 1 mole of $\mathrm{H}_2$ . Therefore, 0.5 moles of Mg will produce 0.5 moles of $\mathrm{H}_2$ .

Step 4: Final Answer: 0.5 moles of $\mathrm{H}_2$

Why other options are wrong:
- Option A: This is incorrect because 12g of Mg is only half a mole, so it cannot produce one full mole of H2.
- Option B: This is incorrect because the reaction produces hydrogen gas (H2), not oxygen gas (O2).
- Option D: This is incorrect because the reaction produces hydrogen gas (H2), not oxygen gas (O2). Also, the amount is wrong.

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

Memory Aids:
- Remember the stoichiometry of the reaction: 1 mole of Mg produces 1 mole of H2.

Quick Tips:
- Always write the balanced chemical equation first.
- Convert mass to moles before using stoichiometry.

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Application +4 | -1

10^{21}

molecules areramoved from

200 {mg}

of

{CO}_2

. The moles of

{CO}_2

left are :

View answer and explanation

Correct option: A

2.84 \times 10^{-3}

The correct answer is A (

2.84 \times 10^{-3}

) because it is the result of subtracting the moles of CO₂ removed from the initial moles of CO₂.

Solution approach

Correct Answer + Core Reason: $2.84 \times 10^{-3}$ . The moles of CO₂ left after removing some molecules.

Step-by-Step Solution:

Step 1: Calculate the number of moles in 200 mg of CO₂: Molar mass of CO₂ = 44 g/mol. Moles = mass / molar mass = (200 mg / 1000 mg/g) / 44 g/mol = 0.2 g / 44 g/mol = $4.5 \times 10^{-3}$ moles.

Step 2: Calculate the number of moles in $10^{21}$ molecules: Moles = Number of molecules / Avogadro's number = $10^{21} / (6.022 \times 10^{23}) = 1.66 \times 10^{-3}$ moles.

Step 3: Calculate the number of moles of CO₂ left: Moles left = Initial moles - Moles removed = $4.5 \times 10^{-3} - 1.66 \times 10^{-3} = 2.84 \times 10^{-3}$ moles.

Why other options are wrong:
- Option B: This is incorrect because it is 10 times larger than the correct answer.
- Option C: This is incorrect because it is 100 times larger than the correct answer.
- Option D: This is incorrect because it is a very large number and has the wrong units.

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

Memory Aids:
- Remember the relationship between moles, mass, and molar mass.

Quick Tips:
- Always convert mass to grams before calculating moles.

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Understanding +4 | -1

For

\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O},

which is the correct mole relationship?

View answer and explanation

Correct option: A

9 \times \text{Mole of }\mathrm{Cu}= \text{Mole of }\mathrm{O}

Option A (

9 \times \text{Mole of }\mathrm{Cu}= \text{Mole of }\mathrm{O}

) is correct because in

\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}

, for every 1 mole of Cu, there are 9 moles of O.

Solution approach

Correct Answer + Core Reason: $9 \times \text{Mole of }\mathrm{Cu}= \text{Mole of }\mathrm{O}$ . The ratio of Cu to O atoms in the compound is 1:9.

Step-by-Step Solution:

Step 1: Analyze the chemical formula: $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ .

Step 2: Determine the number of oxygen atoms per molecule of $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ .
- From $\mathrm{CuSO}_4$ , there are 4 oxygen atoms.
- From $5 \mathrm{H}_2 \mathrm{O}$ , there are 5 oxygen atoms.
- Total oxygen atoms: 4 + 5 = 9.

Step 3: Establish the mole relationship between Cu and O.
- For every 1 mole of $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ , there is 1 mole of Cu and 9 moles of O.
- Therefore, $9 \times \text{Mole of }\mathrm{Cu}= \text{Mole of }\mathrm{O}$ .

Why other options are wrong:
- Option B: This is incorrect because it suggests a 5:1 ratio between Cu and O, which is not true based on the formula.
- Option C: This is incorrect because it relates moles of Cu to moles of $\mathrm{O}_2$ , not moles of O atoms. The formula contains O atoms, not $\mathrm{O}_2$ molecules.
- Option D: This is incorrect because it suggests a 1:5 ratio between Cu and O, which is not true based on the formula.

Key Concepts:
- Mole concept
- Chemical formulas and stoichiometry

Memory Aids:
- Pay close attention to the subscripts in the chemical formula to determine the number of atoms of each element.

Quick Tips:
- Always analyze the chemical formula carefully to determine the correct mole ratios.

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Application +4 | -1

One litre of a gas at STP weight

1.16 \, g

it can possible be

View answer and explanation

Correct option: A

{C}_2 {H}_2

Option A (

{C}_2 {H}_2

) is correct because its molar mass (26 g/mol) is closest to the calculated value based on the given density.

Solution approach

Correct Answer + Core Reason: ${C}_2 {H}_2$ . The molar mass calculated from the given density matches that of acetylene.

Step-by-Step Solution:

Step 1: Calculate the mass of 22.4 L of the gas at STP.

Step 2: Mass of 22.4 L = 22.4 L/mol * 1.16 g/L = 25.984 g/mol ≈ 26 g/mol

Step 3: Compare the calculated molar mass with the molar masses of the given options.

Step 4: Molar mass of ${C}_2 {H}_2$ = 2(12) + 2(1) = 26 g/mol

Why other options are wrong:
- **Option B ( ${CO}$ ): The molar mass of CO is 12 + 16 = 28 g/mol, which is not close to the calculated value.
- Option C ( ${O}_2$ ):** The molar mass of ${O}_2$ is 2(16) = 32 g/mol, which is significantly different from the calculated value.
- **Option D ( ${CH}_4$ ):** The molar mass of ${CH}_4$ is 12 + 4(1) = 16 g/mol, which is also not close to the calculated value.

Key Concepts:
- Molar mass calculation
- STP conditions
- Ideal gas law (implicitly)

Memory Aids:
- Remember that at STP, 1 mole of any gas occupies 22.4 L.

Quick Tips:
- Always convert the given volume to moles using the molar volume at STP.

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Analyzing +4 | -1

Which is correct for

\mathrm{N}{\mathrm{a}_2}\mathrm{HP}{\mathrm{O}_3}

?

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because Na2HPO3 is an acid salt, has an equivalent weight of M/2, and the oxidation number of P is +3.

Solution approach

Correct Answer + Core Reason: All of these. Na2HPO3 is an acid salt, has an equivalent weight of M/2, and the oxidation number of P is +3.

Step-by-Step Solution:

Step 1: Analyze the given compound: Na₂HPO₃

Step 2: Determine if it is an acid salt:
- H₃PO₃ is a dibasic acid (it has two replaceable hydrogen atoms). Therefore, Na₂HPO₃ is a normal salt.
- Since it is derived from a dibasic acid and has two sodium atoms, it is considered a normal salt.

Step 3: Calculate the equivalent weight:
- The basicity of H₃PO₃ is 2. Therefore, the equivalent weight of Na₂HPO₃ is M/2, where M is the molar mass.

Step 4: Determine the oxidation number of P:
- Let the oxidation number of P be x.
- The oxidation numbers of Na, H, and O are +1, +1, and -2, respectively.
- The sum of the oxidation numbers in Na₂HPO₃ is 0.
- 2(+1) + (+1) + x + 3(-2) = 0
- 2 + 1 + x - 6 = 0
- x - 3 = 0
- x = +3

Step 5: Evaluate the given options:
- Option A: It is not an acid salt. This is correct because Na₂HPO₃ is a normal salt.
- Option B: Eq. wt. = M/2. This is correct because the basicity of H₃PO₃ is 2.
- Option C: Ox. no. of P is +3. This is correct as calculated above.

Why other options are wrong:
- Option A (It is not an acid salt): This statement is correct, but it's not the only correct statement. Since all the statements are correct, option D is the correct answer.
- Option B (Eq. wt. = M/2): This statement is correct, but it's not the only correct statement. Since all the statements are correct, option D is the correct answer.
- Option C (Ox. no. of P is + 3): This statement is correct, but it's not the only correct statement. Since all the statements are correct, option D is the correct answer.

Key Concepts:
- Acid salts and normal salts
- Basicity of acids
- Equivalent weight
- Oxidation number

Memory Aids:
- Remember that H₃PO₃ is a dibasic acid.

Quick Tips:
- Determine the basicity of the acid to calculate the equivalent weight of the salt.
- Use the oxidation number rules to calculate the oxidation number of an element in a compound.

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Application +4 | -1

The mass of 2.5 g.atom of oxygen is

View answer and explanation

Correct option: A

40 g

Option A (40 g) is correct because 1 g.atom is equal to the gram atomic weight, and for oxygen, it's 16 g/g.atom. Therefore, 2.5 g.atom is 2.5 * 16 = 40 g.

Solution approach

Correct Answer + Core Reason: 40 g. The mass is calculated by multiplying the number of gram atoms by the gram atomic weight of oxygen.

Step-by-Step Solution:

Step 1: Identify the given information: 2.5 g.atom of oxygen.

Step 2: Recall that 1 g.atom of an element is equal to its gram atomic weight.

Step 3: The atomic weight of oxygen (O) is approximately 16 g/mol.

Step 4: Calculate the mass: Mass = (number of g.atoms) * (atomic weight)\nMass = 2.5 g.atom * 16 g/g.atom

Step 5: Mass = 40 g

Why other options are wrong:
- Option B (4 g): This is incorrect. It might be a misunderstanding of the concept of gram atoms.
- Option C (0.4): This is incorrect and doesn't relate to the calculation.
- Option D (400): This is incorrect. It might be a calculation error or misunderstanding of the concept.

Key Concepts:
- Gram atomic weight
- Moles

Memory Aids:
- 1 g.atom = gram atomic weight

Quick Tips:
- Always remember the definition of gram atomic weight.

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Analyzing +4 | -1

Statement - I: For reaction:

2A(g) + 3B(g) \rightarrow 4C(g) + D(g),

the vapour density remain
constant throughout the progress of reaction.

Statement - II: In all the gaseous chemical reactions, the vapour density remains constant.

View answer and explanation

Correct option: C

If Statement-I is true but Statement-II is false

Option C is correct because Statement-I is true (moles remain constant, so vapor density remains constant), but Statement-II is false (vapor density doesn't always remain constant in gaseous reactions).

Solution approach

Correct Answer + Core Reason: If Statement-I is true but Statement-II is false

Step-by-Step Solution:

Step 1: Analyze Statement-I. For the reaction $2A(g) + 3B(g) \rightarrow 4C(g) + D(g)$ , the total number of moles of gas is 2 + 3 = 5 on the reactant side and 4 + 1 = 5 on the product side. Since the number of moles remains constant, the vapor density remains constant as well.

Step 2: Analyze Statement-II. Statement-II claims that in all gaseous chemical reactions, the vapor density remains constant. This is not true. Vapor density depends on the number of moles of gas. If the number of moles changes during the reaction, the vapor density will also change.

Step 3: Combine the analysis. Statement-I is true, but Statement-II is false.

Why other options are wrong:
- Option A: Incorrect because Statement-II is false.
- Option B: Incorrect because Statement-II is false.
- Option D: Incorrect because Statement-I is true.

Key Concepts:
- Vapor density
- Stoichiometry
- Gaseous reactions

Memory Aids:
- Vapor density is related to the molar mass and the number of moles of gas.

Quick Tips:
- Check if the number of moles of gas changes during the reaction to determine if the vapor density changes.

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Understanding +4 | -1

Assertion (A): A one molal solution prepared at

20^{\circ} \mathrm{C}

will retain the same molality at

100^{\circ} \mathrm{C},

provided there is no loss of solute or solvent on heating.

Reason (R): Molality is independent of temperature.

View answer and explanation

Correct option: A

Both (A) & (R) are true and the (R) is the correct explanation of the (A)

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because molality is defined as moles of solute per kg of solvent, which is independent of temperature, and this explains why molality remains constant with temperature changes.

Solution approach

Correct Answer + Core Reason: Both (A) & (R) are true and the (R) is the correct explanation of the (A). Molality is temperature independent.

Step-by-Step Solution:

Step 1: Define Molality: Molality (m) is defined as the number of moles of solute per kilogram of solvent: $m = \frac{{\text{{moles of solute}}}}{{\text{{kg of solvent}}}}$

Step 2: Temperature Dependence: Since molality is based on mass, which does not change with temperature, molality is independent of temperature.

Step 3: Analyze Assertion: A one molal solution at 20°C will remain one molal at 100°C if there is no loss of solute or solvent.

Step 4: Analyze Reason: Molality is independent of temperature.

Step 5: Conclusion: Both the assertion and the reason are true, and the reason correctly explains the assertion.

Why other options are wrong:
- Option A: Both (A) & (R) are true and the (R) is the correct explanation of the (A) - Correct because molality is temperature independent.
- Option B: Both (A) & (R) are true but the (R) is not the correct explanation of the (A) - Incorrect because the reason directly explains the assertion.
- Option C: (A) is true but (R) is false - Incorrect because both are true.
- Option D: Both (A) and (R) are false - Incorrect because both are true.

Key Concepts:
- Molality
- Temperature Dependence of Concentration

Memory Aids:
- Remember that molality is based on mass, not volume, making it temperature independent.

Quick Tips:
- Contrast molality with molarity, which is temperature-dependent due to volume changes.

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Application +4 | -1

A nitrogen containing organic compound contain 28.9% nitrogen by weight. If one molecule compound have 4 nitrogen atom then molecular weight of compound is:

View answer and explanation

Correct option: B

194

Option B (194) is correct because it is the molecular weight calculated using the percentage of nitrogen and the number of nitrogen atoms in the molecule.

Solution approach

Correct Answer + Core Reason: 194. The molecular weight is calculated using the percentage of nitrogen by weight and the number of nitrogen atoms.

Step-by-Step Solution:
Step 1: Define the formula for percentage of nitrogen: $\text{% of N} = \frac{{\text{{no. of atoms}} \times \text{{G.A.M}}}}{{\text{{G.M.M}}}} \times 100$
Step 2: Substitute the given values: $28.9 = \frac{{4 \times 14}}{x} \times 100$
Step 3: Solve for x (molecular weight): $x = \frac{{4 \times 14 \times 100}}{{28.9}} = 193.77 \approx 194$

Why other options are wrong:
- Option A (164): This value is incorrect because it doesn't correctly account for the percentage of nitrogen.
- Option C (254): This value is incorrect because it doesn't correctly account for the percentage of nitrogen.
- Option D (108): This value is incorrect because it doesn't correctly account for the percentage of nitrogen.

Key Concepts:
- Percentage composition
- Molecular weight calculation

Memory Aids:
- Remember the formula: Percentage = (Part/Whole) * 100

Quick Tips:
- Always check the units and make sure the formula is correctly applied.

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Application +4 | -1

0.72 \mathrm{~gm}

of an oxide of a metal

\mathrm{M}

on reduction with

\mathrm{H}_2

gave

0.64 \mathrm{~g}

of the metal. The atomic weight of the metal is 64. The empirical formula of the compound is

View answer and explanation

Correct option: B

\mathrm{M}_2 \mathrm{O}

Option B (M₂O) is correct because the empirical formula is determined by calculating the mole ratio of the metal and oxygen, which simplifies to 2:1.

Solution approach

Correct Answer + Core Reason: $\mathrm{M}_2 \mathrm{O}$ . The empirical formula is determined by finding the mole ratio of metal to oxygen.

Step-by-Step Solution:

Step 1: Calculate the mass of oxygen: Mass of oxygen = Mass of oxide - Mass of metal = 0.72 g - 0.64 g = 0.08 g

Step 2: Calculate the moles of metal: Moles of metal = $\frac{\text{{Mass of metal}}}{\text{{Atomic weight of metal}}} = \frac{0.64 \text{{ g}}}{64 \text{{ g/mol}}} = 0.01 \text{{ mol}}$

Step 3: Calculate the moles of oxygen: Moles of oxygen = $\frac{\text{{Mass of oxygen}}}{\text{{Atomic weight of oxygen}}} = \frac{0.08 \text{{ g}}}{16 \text{{ g/mol}}} = 0.005 \text{{ mol}}$

Step 4: Determine the mole ratio of metal to oxygen: $\frac{\text{{Moles of metal}}}{\text{{Moles of oxygen}}} = \frac{0.01}{0.005} = 2$

Step 5: Write the empirical formula: Since the mole ratio of M to O is 2:1, the empirical formula is $\mathrm{M}_2 \mathrm{O}$

Why other options are wrong:
- **Option A ( $\mathrm{MO}$ ): This is incorrect because the mole ratio of metal to oxygen is not 1:1.
- Option C ( $\mathrm{MO}_2$ ): This is incorrect because the mole ratio of metal to oxygen is not 1:2.
- Option D ( $\mathrm{M}_2 \mathrm{O}_3$ ): This is incorrect because the mole ratio of metal to oxygen is not 2:3.

Key Concepts:
- Empirical formula
- Mole concept
- Stoichiometry

Memory Aids:
- Remember that the empirical formula represents the simplest whole-number ratio of atoms in a compound.

Quick Tips:**
- Always convert masses to moles before determining the empirical formula.
- Simplify the mole ratio to the smallest whole numbers.

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Understanding +4 | -1

Molecular weight of tribasic acid is W. Its equivalent weight will be:

View answer and explanation

Correct option: B

\frac{W}{3}

Option B (

\frac{W}{3}

) is correct because equivalent weight is calculated by dividing the molecular weight by the basicity of the acid.

Solution approach

Correct Answer + Core Reason: $\frac{W}{3}$ . Equivalent weight is molecular weight divided by basicity.

Step-by-Step Solution:

Step 1: Recall the formula for equivalent weight. Equivalent weight = (Molecular weight) / (Basicity or Acidity or Number of electrons transferred).

Step 2: Apply the formula to the given problem. The molecular weight is W, and the acid is tribasic, so the basicity is 3.

Step 3: Calculate the equivalent weight. Equivalent weight = W / 3.

Why other options are wrong:
- **Option A ( $\frac{W}{2}$ ): Incorrect. This would be the equivalent weight of a dibasic acid.
- Option C ( $W$ ): Incorrect. This is the molecular weight, not the equivalent weight.
- Option D ( $3W$ ): Incorrect. This is the molecular weight multiplied by the basicity, which is not the equivalent weight.

Key Concepts:
- Equivalent weight
- Basicity of an acid

Memory Aids:
- Remember that equivalent weight = (Molecular weight) / (Basicity).

Quick Tips:**
- Know the formula for calculating equivalent weight.

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Understanding +4 | -1

Which of the following statements is correct about the reaction given below?

4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~g})

View answer and explanation

Correct option: A

Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore, it follows law of conservation of mass.

Option A is correct because the total mass of reactants equals the total mass of products, illustrating the law of conservation of mass.

Solution approach

Correct Answer + Core Reason: Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore, it follows law of conservation of mass. Mass is conserved in a chemical reaction.

Step-by-Step Solution:

Step 1: State the Law of Conservation of Mass: The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction.

Step 2: Analyze the Reaction: $4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_2(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{g})$

Step 3: Calculate Molar Masses: Molar mass of Fe = 56 g/mol, Molar mass of O2 = 32 g/mol, Molar mass of Fe2O3 = 160 g/mol

Step 4: Calculate Total Mass of Reactants: (4 * 56) + (3 * 32) = 224 + 96 = 320 g

Step 5: Calculate Total Mass of Products: 2 * 160 = 320 g

Step 6: Conclusion: The total mass of reactants equals the total mass of products, demonstrating the law of conservation of mass.

Why other options are wrong:
- Option B: Law of multiple proportions applies to different compounds formed from the same elements, not a single reaction.
- Option C: The amount of product is limited by the limiting reactant, not by reducing any reactant.
- Option D: Excess of one reactant does not decrease the product formed; it's determined by the limiting reactant.

Key Concepts:
- Law of Conservation of Mass
- Stoichiometry
- Limiting Reactant

Memory Aids:
- Mass in = Mass out (in a chemical reaction).

Quick Tips:
- Balance the chemical equation before analyzing mass relationships.

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Application +4 | -1

Number of atoms in 558.5 g of Fe is (at . wt = 55.85)

View answer and explanation

Correct option: A

Twice that in 60 g carbon

Option A (Twice that in 60 g carbon) is correct because 558.5 g of Fe is 10 moles, and 60 g of carbon is 5 moles, so the number of atoms in Fe is twice that in carbon.

Solution approach

Correct Answer + Core Reason: Twice that in 60 g carbon. The number of atoms is proportional to the number of moles.

Step-by-Step Solution:

Step 1: Calculate the number of moles of Fe: Moles of Fe = $\frac{{\text{{given mass}}}}{{\text{{atomic weight}}}} = \frac{{558.5 \text{{ g}}}}{{55.85 \text{{ g/mol}}}} = 10 \text{{ mol}}$

Step 2: Calculate the number of atoms of Fe: Number of atoms of Fe = Moles of Fe $\times$ Avogadro's number = $10 \times N_A$ , where $N_A$ is Avogadro's number.

Step 3: Calculate the number of moles of C: Moles of C = $\frac{{\text{{given mass}}}}{{\text{{atomic weight}}}} = \frac{{60 \text{{ g}}}}{{12 \text{{ g/mol}}}} = 5 \text{{ mol}}$

Step 4: Calculate the number of atoms of C: Number of atoms of C = Moles of C $\times$ Avogadro's number = $5 \times N_A$

Step 5: Compare the number of atoms of Fe and C: $\frac{{\text{{Number of atoms of Fe}}}}{{\text{{Number of atoms of C}}}} = \frac{{10 \times N_A}}{{5 \times N_A}} = 2$ \nTherefore, the number of atoms in 558.5 g of Fe is twice that in 60 g of carbon.

Step 6: Calculate the number of moles of He: Moles of He = $\frac{{\text{{given mass}}}}{{\text{{atomic weight}}}} = \frac{{8 \text{{ g}}}}{{4 \text{{ g/mol}}}} = 2 \text{{ mol}}$

Step 7: Calculate the number of atoms of He: Number of atoms of He = Moles of He $\times$ Avogadro's number = $2 \times N_A$

Why other options are wrong:
- **Option B ( $6.023 \times {10^{22}}$ ): This is incorrect because it's much smaller than the actual number of atoms in 10 moles of Fe.
- Option C (Half in 8 gr of He): This is incorrect because 558.5 g of Fe contains 10 moles of atoms, while 8 g of He contains 2 moles of atoms. So, Fe has 5 times the number of atoms in 8 g of He, not half.
- Option D ( $558.5 \times 6.023 \times {10^{23}}$ ): This is incorrect because it multiplies the mass by Avogadro's number, which is not the correct way to calculate the number of atoms.

Key Concepts:
- Moles
- Avogadro's number
- Atomic weight

Memory Aids:**
- Number of atoms = Moles $\times$ Avogadro's number

Quick Tips:
- Always convert mass to moles before calculating the number of atoms.
- Remember that the number of atoms is directly proportional to the number of moles.

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Analyzing +4 | -1

Assertion: Density is expressed as

\mathrm{g} \mathrm{~mL}^{-1}

whereas specific gravity is dimensionless.
Reason: Specific gravity is ratio of the masses of solution and solvent.

View answer and explanation

Correct option: C

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

The correct answer is C because specific gravity is the ratio of densities (or masses) of a substance to a reference substance, making it dimensionless, and the reason correctly explains this.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct and Reason is correct explanation of Assertion. Density has units, while specific gravity is a ratio and therefore dimensionless.

Step-by-Step Solution:
Step 1: Define Density: Density is mass per unit volume, typically expressed in $\mathrm{g} \mathrm{~mL}^{-1}$ or $\mathrm{g} \mathrm{~cm}^{-3}$ .
Step 2: Define Specific Gravity: Specific gravity is the ratio of the density of a substance to the density of a reference substance (usually water for liquids and solids, and air for gases).
Step 3: Analyze the Assertion: Density is expressed as $\mathrm{g} \mathrm{~mL}^{-1}$ , which is correct. Specific gravity is dimensionless because it is a ratio of two densities, so the units cancel out. The assertion is correct.
Step 4: Analyze the Reason: Specific gravity is the ratio of the masses of solution and solvent. This is a simplified view. More accurately, it's the ratio of the density of a substance to the density of a reference substance. However, since density is directly proportional to mass (at constant volume), the statement is essentially correct in the context of specific gravity being a ratio.
Step 5: Determine if the Reason explains the Assertion: The reason explains why specific gravity is dimensionless because it is a ratio of two quantities with the same units, which cancel out.

Why other options are wrong:
- Option A: The reason is not wrong, it correctly explains why specific gravity is dimensionless.
- Option B: The assertion is correct, not wrong.
- Option D: The reason is a correct explanation of the assertion.

Key Concepts:
- Density
- Specific Gravity
- Dimensional Analysis

Memory Aids:
- Specific gravity is a *ratio*, so it has no units.

Quick Tips:
- Remember that specific gravity is a comparison to a reference substance, usually water.

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Application +4 | -1

Sodium bicarbonate

\mathrm{NaHCO}_3

can be purified by dissolving it in hot water

\left(60^{\circ} \mathrm{C}\right),

filtering to remove insoluble impurities, cooling to

0^{\circ} \mathrm{C}

to precipitate solid

\mathrm{NaHCO}_3,

and then filtering to remove the solid, leaving soluble impurities in solution. Any

\mathrm{NaHCO}_3

that remains in the solution is not recovered. The solubility of

\mathrm{NaHCO}_3

in hot water at

60^{\circ} \mathrm{C}

is

164 \mathrm{~g} /

litre and is

69 \mathrm{~g/litre}

in cold water at

0^{\circ} \mathrm{C}.

What is the percentage yield of

\mathrm{NaHCO}_3,

when it is purified by this method?

View answer and explanation

Correct option: A

57.93 \%

Option A (

57.93 \%

) is correct because the percentage yield is calculated by finding the difference in solubility between hot and cold water, then dividing by the solubility in hot water.

Solution approach

Correct Answer + Core Reason: $57.93 \%$ . The percentage yield is calculated by finding the difference in solubility between hot and cold water, then dividing by the solubility in hot water.

Step-by-Step Solution:

Step 1: Determine the amount of NaHCO3 that precipitates out: Solubility at 60°C - Solubility at 0°C = 164 g/L - 69 g/L = 95 g/L

Step 2: Calculate the percentage yield: (Amount precipitated / Initial amount) * 100 = (95 g/L / 164 g/L) * 100

Step 3: Calculate the percentage: (95 / 164) * 100 = 0.579268 * 100 = 57.93%

Step 4: Final Answer: $57.93 \%$

Why other options are wrong:
- Option B: $42.07 \%$ : This is the percentage of NaHCO3 that remains dissolved, not the yield.
- Option C: $69 \%$ : This is the solubility in cold water, not the yield.
- Option D: $100 \%$ : This would only be true if all the NaHCO3 precipitated out, which is not the case.

Key Concepts:
- Solubility
- Percentage Yield

Memory Aids:
- Percentage Yield = (Actual Yield / Theoretical Yield) * 100

Quick Tips:
- Always subtract the final amount from the initial amount to find the amount precipitated.

180 Medium

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Application +4 | -1

\mathrm{NaClO}

solution reacts with

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3},as,\mathrm{NaClO} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3} \to \mathrm{NaCl} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

A solution of

\mathrm{NaClO}

used in the above reaction contained 15 g of

\mathrm{NaClO}

per litre. The normality of the solution would be :

View answer and explanation

Correct option: A

0.4

Option A (0.4) is correct because the normality is calculated by considering the number of equivalents of NaClO per liter of solution.

Solution approach

Correct Answer + Core Reason: 0.4. The normality is calculated by considering the number of equivalents of $\mathrm{NaClO}$ per liter of solution.

Step-by-Step Solution:

Step 1: Write the reaction: $\mathrm{NaClO} + {\mathrm{H}_2}\mathrm{SO}_3 \to \mathrm{NaCl} + {\mathrm{H}_2}\mathrm{SO}_4$

Step 2: Determine the change in oxidation state of chlorine. In $\mathrm{NaClO}$ , the oxidation state of Cl is +1. In $\mathrm{NaCl}$ , the oxidation state of Cl is -1. The change is 2.

Step 3: Calculate the equivalent weight of $\mathrm{NaClO}$ : Equivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{Change in oxidation state}}}} = \frac{{74.5}}{2} = 37.25$

Step 4: Calculate the normality: Normality = $\frac{{\text{{Grams per liter}}}}{{\text{{Equivalent weight}}}} = \frac{{15}}{{37.25}} = 0.40$

Why other options are wrong:
- Option B (0.2): Incorrect because it uses a different equivalent weight or an incorrect calculation.
- Option C (0.6): Incorrect because it uses a different equivalent weight or an incorrect calculation.
- Option D (0.8): Incorrect because it uses a different equivalent weight or an incorrect calculation.

Key Concepts:
- Normality
- Equivalent weight
- Oxidation state

Memory Aids:
- Normality = Molarity x n-factor

Quick Tips:
- Remember to determine the change in oxidation state to find the n-factor.
- Pay attention to the units and conversions.

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