SOME BASIC CONCEPTS OF CHEMISTRY MCQs for JEE CHEMISTRY — Medium, Page 3

61 Medium

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Application +4 | -1

If potassium chlorate is

80 \%

pure, then

48 \mathrm{~g}

of oxygen would be produced from (atomic mass of

\mathrm{K}=39

)

View answer and explanation

Correct option: A

153.12 \mathrm{~g} \mathrm{of} \mathrm{KClO}_3

Option A (

153.12 \mathrm{~g} \mathrm{of} \mathrm{KClO}_3

) is correct because it accounts for the purity of potassium chlorate to produce 48g of oxygen.

Solution approach

Correct Answer + Core Reason: $153.12 \mathrm{~g} \mathrm{of} \mathrm{KClO}_3$ . The mass of $KClO_3$ needed is calculated considering its purity.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the decomposition of potassium chlorate:
$2KClO_3 \xrightarrow{{\text{heat}}} 2KCl + 3O_2$

Step 2: Determine the molar masses of $KClO_3$ and $O_2$ :\nMolar mass of $KClO_3 = 39 + 35.5 + 3(16) = 122.5 \text{ g/mol}$ \nMolar mass of $O_2 = 2(16) = 32 \text{ g/mol}$

Step 3: Calculate the mass of $KClO_3$ required to produce 48 g of $O_2$ based on stoichiometry:\nFrom the balanced equation, 2 moles of $KClO_3$ produce 3 moles of $O_2$ .
$2 \times 122.5 \text{ g } KClO_3 \rightarrow 3 \times 32 \text{ g } O_2$
$245 \text{ g } KClO_3 \rightarrow 96 \text{ g } O_2$ \nSo, to produce 48 g of $O_2$ , we need:
$\frac{245 \text{ g } KClO_3}{96 \text{ g } O_2} \times 48 \text{ g } O_2 = 122.5 \text{ g } KClO_3$

Step 4: Account for the purity of $KClO_3$ :\nSince the $KClO_3$ is only 80% pure, we need to calculate the actual amount of impure $KClO_3$ required:
$\text{Amount of impure } KClO_3 = \frac{100}{80} \times 122.5 \text{ g} = 153.125 \text{ g}$

Step 5: Round to two decimal places.
$153.12 \text{ g}$

Why other options are wrong:
- **Option B ( $245 \mathrm{~g}$ of $\mathrm{KClO}_3$ ):** This is the mass of $KClO_3$ if it were 100% pure and produced 96g of oxygen. It doesn't account for the 80% purity or the 48g of oxygen.
- **Option C ( $122.5 \mathrm{~g}$ of $\mathrm{KClO}_3$ ):** This is the mass of pure $KClO_3$ needed to produce 48g of oxygen, but it doesn't account for the fact that the $KClO_3$ is only 80% pure.
- **Option D ( $98 \mathrm{~g}$ of $\mathrm{KClO}_3$ ): This value is incorrect and doesn't relate to the stoichiometry or purity of the reaction.

Key Concepts:
- Stoichiometry
- Percentage purity
- Molar mass

Memory Aids:
- Remember to always balance the chemical equation before performing stoichiometric calculations.
- Purity = (Mass of pure substance / Mass of impure substance) x 100

Quick Tips:**
- Always check if the given substance is pure or impure. If impure, adjust the calculations accordingly.

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Application +4 | -1

The density of

3 \mathrm{~M}

solution of

\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3

is

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

Calculate, the mole fraction of

\mathrm{Na}_2 \mathrm{SO}_3.

View answer and explanation

Correct option: A

0.065

Option A (0.065) is correct because the mole fraction is calculated by dividing moles of Na2S2O3 by the total moles of solute and solvent.

Solution approach

Correct Answer + Core Reason: 0.065. The mole fraction is calculated by dividing the moles of solute by the total moles in the solution.

Step-by-Step Solution:

Step 1: Given molarity of $Na_2S_2O_3$ is 3 M, which means 3 moles of $Na_2S_2O_3$ are present in 1 L of solution.

Step 2: Calculate the weight of $Na_2S_2O_3$ : $3 \times 158 = 474 \text{{ g}}$ (Molar mass of $Na_2S_2O_3 = 158 \text{{ g/mol}}$ )

Step 3: Volume of solution = 1 L = 1000 mL.

Step 4: Weight of solution = Volume × Density = $1000 \times 1.25 = 1250 \text{{ g}}$

Step 5: Weight of water = Weight of solution - Weight of $Na_2S_2O_3 = 1250 - 474 = 776 \text{{ g}}$

Step 6: Moles of water = Weight of water / Molar mass of water = $776 / 18 = 43.11$

Step 7: Mole fraction of $Na_2S_2O_3 = \frac{{\text{{Moles of }} Na_2S_2O_3}}{{\text{{Moles of }} Na_2S_2O_3 + \text{{Moles of }} H_2O}} = \frac{{3}}{{3 + 43.11}} = \frac{{3}}{{46.11}} = 0.065$

Why other options are wrong:
- Option B (0.13): This is incorrect. It might be calculated by not considering the correct molar mass or by making errors in the mole fraction calculation.
- Option C (0.26): This is incorrect. It might be calculated by not considering the correct molar mass or by making errors in the mole fraction calculation.
- Option D (0.39): This is incorrect. It might be calculated by not considering the correct molar mass or by making errors in the mole fraction calculation.

Key Concepts:
- Molarity
- Mole fraction
- Density

Memory Aids:
- Mole fraction = Moles of component / Total moles

Quick Tips:
- Always convert mass to moles before calculating mole fraction.
- Pay attention to units and molar masses.

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Application +4 | -1

The number of electron in

3.1 \mathrm{~mg} \mathrm{~NO}_3^{-}

is:-

View answer and explanation

Correct option: C

9.6 \times 10^{20}

Option C (9.6 × 10²⁰) is correct because the number of electrons is calculated by finding the moles of NO₃⁻, then multiplying by Avogadro's number and the number of electrons in the ion.

Solution approach

Correct Answer + Core Reason: $9.6 \times 10^{20}$ . The number of electrons is calculated based on the number of moles of nitrate ions and the number of electrons per ion.

Step-by-Step Solution:

Step 1: Calculate the molar mass of NO₃⁻:\nMolar mass (NO₃⁻) = 14 (N) + 3 * 16 (O) = 14 + 48 = 62 g/mol

Step 2: Calculate the number of moles of NO₃⁻:\nMoles (NO₃⁻) = mass / molar mass = (3.1 mg) / (62 g/mol) = (3.1 * 10⁻³ g) / (62 g/mol) = 0.05 * 10⁻³ mol = 5 * 10⁻⁵ mol

Step 3: Calculate the number of NO₃⁻ ions:\nNumber of ions = moles * Avogadro's number = (5 * 10⁻⁵ mol) * (6.022 * 10²³ ions/mol) = 3.011 * 10¹⁹ ions

Step 4: Calculate the number of electrons in one NO₃⁻ ion:\nElectrons in N = 7\nElectrons in 3 O = 3 * 8 = 24\nCharge = -1, so add 1 electron\nTotal electrons = 7 + 24 + 1 = 32 electrons/ion

Step 5: Calculate the total number of electrons:\nTotal electrons = Number of ions * electrons per ion = (3.011 * 10¹⁹ ions) * (32 electrons/ion) = 96.352 * 10¹⁹ electrons ≈ 9.6 * 10²⁰ electrons

Why other options are wrong:
- Option A (32): This is incorrect. It only represents the number of electrons in a single nitrate ion, not the total number of electrons in the given mass.
- **Option B ( $1.6 \times 10^{-3}$ ): This is incorrect. There is no clear logical error that leads to this answer.
- Option D ( $9.6 \times 10^{23}$ ): This is incorrect. It might be a calculation error in multiplying by Avogadro's number.

Key Concepts:
- Molar mass
- Moles
- Avogadro's number

Memory Aids:**
- Moles = mass / molar mass
- Number of particles = moles * Avogadro's number

Quick Tips:
- Always pay attention to units and convert them to be consistent.

64 Medium

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Analyzing +4 | -1

\mathrm{N}_2

and

\mathrm{H}_2

are mixed in

14:3

mass ratio. After certain time ammonia was found to be

40 \%

by mol. The mole fraction of

\text{N}_2

at that time in the mixture of

\mathrm{N}_2,\mathrm{H}_2

and

\text{NH}_3

is

View answer and explanation

Correct option: A

0.15

Option A (0.15) is correct because after calculating the moles of each component at equilibrium, the mole fraction of N2 is found to be 0.15.

Solution approach

Correct Answer + Core Reason: 0.15. The mole fraction of N2 is calculated based on the equilibrium moles of N2, H2, and NH3.

Step-by-Step Solution:

Step 1: Calculate initial moles of N2 and H2.\nMoles of N2 = $\frac{{14x}}{{28}} = \frac{{x}}{{2}}$ \nMoles of H2 = $\frac{{3x}}{{2}}$

Step 2: Write the balanced reaction and set up an ICE table:
${{\text{N}}_2} + {\text{3}}{{\text{H}}_2} \to 2{\text{N}}{{\text{H}}_3}$ \nInitial: $\frac{{x}}{{2}}$ $\frac{{3x}}{{2}}$ 0\nChange: -a -3a +2a\nEquilibrium: $\frac{{x}}{{2}} - a$ $\frac{{3x}}{{2}} - 3a$ 2a

Step 3: Use the given information that NH3 is 40% by mole fraction:
$\frac{{2a}}{{\left( {\frac{{x}}{{2}} - a} \right) + \left( {\frac{{3x}}{{2}} - 3a} \right) + 2a}} = 0.4$
$\frac{{2a}}{{2x - 2a}} = 0.4$
$\frac{{a}}{{x - a}} = 0.4$
$a = 0.4x - 0.4a$
$1.4a = 0.4x$
$\frac{{x}}{{a}} = \frac{{1.4}}{{0.4}} = \frac{{7}}{{2}}$

Step 4: Calculate the mole fraction of N2:
$x_{N_2} = \frac{{\frac{{x}}{{2}} - a}}{{2x - 2a}} = \frac{{\frac{{x}}{{2}} - a}}{{2(x - a)}} = \frac{{\frac{{x}}{{2a}} - 1}}{{2(\frac{{x}}{{a}} - 1)}} = \frac{{\frac{{7}}{{4}} - 1}}{{2(\frac{{7}}{{2}} - 1)}} = \frac{{\frac{{3}}{{4}}}}{{2(\frac{{5}}{{2}})}} = \frac{{\frac{{3}}{{4}}}}{5} = \frac{{3}}{{20}} = 0.15$

Why other options are wrong:
- Option B (0.3): This value is incorrect because it doesn't properly account for the stoichiometry of the reaction and the equilibrium mole fractions.
- Option C (0.45): This value is incorrect because it doesn't properly account for the stoichiometry of the reaction and the equilibrium mole fractions.
- Option D (None of these): This is incorrect as 0.15 is the correct answer.

Key Concepts:
- Mole fraction
- Stoichiometry
- Equilibrium

Memory Aids:
- ICE table for equilibrium calculations

Quick Tips:
- Always balance the chemical equation first.
- Use an ICE table to track changes in moles.

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Understanding +4 | -1

X

' is the amount of a substance that contains as many particles as there are atoms in exactly

12 \mathrm{~g}

of the

Y

isotope.

X

and

Y

respectively are

View answer and explanation

Correct option: C

A \rightarrow

one mole,

B \rightarrow{ }^{12} \mathrm{C}

Option C is correct because one mole is defined as the amount of substance containing as many particles as there are atoms in exactly 12g of the carbon-12 isotope.

Solution approach

Correct Answer + Core Reason: $A \rightarrow$ one mole, $B \rightarrow{ }^{12} \mathrm{C}$ . This is the definition of a mole.

Step-by-Step Solution:
Step 1: Recall the definition of a mole.
Step 2: The mole is defined as the amount of substance containing as many particles as there are atoms in exactly 12 g of the $^{12}C$ isotope.
Step 3: One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly $12 \mathrm{~g}\,(0.012 \mathrm{~kg})$ of the ${ }^{12} \mathrm{C}$ isotope.
Step 4: Thus, $X$ and $Y$ respectively are one mole and ${ }^{12} \mathrm{C}$ .

Why other options are wrong:
- Option A: $^{14}C$ is not used in the definition of a mole.
- Option B: Percentage composition is not related to the definition of a mole.
- Option D: Percentage composition is not related to the definition of a mole, and $^{14}C$ is not used in the definition of a mole.

Key Concepts:
- Mole concept
- Avogadro's number
- Isotopes

Memory Aids:
- Remember the definition of a mole in terms of $^{12}C$ .

Quick Tips:
- Understand the fundamental definitions in chemistry.

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Application +4 | -1

A mixture contains

{Na}_2 {C}_2 {O}_4

and

{KHC}_2 {O}_4

in 1:1 molar ratio. Mixture is neutralised by

100 \mathrm{~mL}

of

0.01 \, \mathrm{M} \, {KOH}

. What volume of

0.01 \, \mathrm{M} \, {KMnO}_4?

Thus, the same mixture is oxidised by

View answer and explanation

Correct option: D

80 \mathrm{~mL}

Option D (

80 \mathrm{~mL}

) is correct because it is calculated by finding the millimoles of

{C}_2 {O}_4^{2-}

and equating it to the milliequivalents of

KMnO_4

.

Solution approach

Correct Answer + Core Reason: $80 \mathrm{~mL}$ . The volume of $0.01 \, \mathrm{M} \, {KMnO}_4$ is calculated based on the stoichiometry of the reaction with oxalate ions.

Step-by-Step Solution:

Step 1: Understand the neutralization reaction:
${KHC}_2 {O}_4$ reacts with ${KOH}$ in a 1:1 molar ratio.

Step 2: Calculate millimoles of ${KOH}$ used for neutralization:\nMillimoles of ${KOH} = 100 \mathrm{~mL} \times 0.01 \mathrm{~M} = 1 \text{ millimol}$

Step 3: Determine millimoles of ${KHC}_2 {O}_4$ :\nSince ${KHC}_2 {O}_4 \equiv {KOH}$ , millimoles of ${KHC}_2 {O}_4 = 1 \text{ millimol}$

Step 4: Determine millimoles of ${Na}_2 {C}_2 {O}_4$ :\nThe mixture contains ${Na}_2 {C}_2 {O}_4$ and ${KHC}_2 {O}_4$ in a 1:1 molar ratio. Therefore, millimoles of ${Na}_2 {C}_2 {O}_4 = 1 \text{ millimol}$

Step 5: Calculate total millimoles of ${C}_2 {O}_4^{2-}$ :\nTotal millimoles of ${C}_2 {O}_4^{2-} = 1 \text{ millimol (from } {Na}_2 {C}_2 {O}_4) + 1 \text{ millimol (from } {KHC}_2 {O}_4) = 2 \text{ millimol}$

Step 6: Determine milliequivalents of ${C}_2 {O}_4^{2-}$ :
${C}_2 {O}_4^{2-} \rightarrow 2 {CO}_2 + 2e^-$ , so n-factor = 2\nMilliequivalents of ${C}_2 {O}_4^{2-} = 2 \text{ millimol} \times 2 = 4 \text{ milliequiv}$

Step 7: Determine the n-factor for ${KMnO}_4$ :
${MnO}_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ , so n-factor = 5

Step 8: Let V be the volume of $0.01 \, \mathrm{M} \, {KMnO}_4$ in mL. Then, milliequivalents of ${KMnO}_4 = V \times 0.01 \times 5$

Step 9: Equate milliequivalents of ${KMnO}_4$ and ${C}_2 {O}_4^{2-}$ :\nV × 0.01 × 5 = 4

Step 10: Solve for V:\nV = $\frac{4}{0.01 × 5}$ = $\frac{4}{0.05}$ = 80 ~mL $\n\n**Step 11:** Final Answer:$ 80 ~mL $\n\n**Why other options are wrong:**\n- **Option A ($ 200 ~mL $):** This is incorrect due to an error in the calculation, possibly not considering the n-factor of both$ {KMnO}_4 $and$ {C}_2 {O}_4^2- $.\n- **Option B ($ 100 ~mL $):** This is incorrect due to an error in the calculation, possibly not considering the n-factor of$ {KMnO}_4 $.\n- **Option C ($ 90 ~mL $):** This is incorrect due to an error in the calculation, possibly using incorrect stoichiometry.\n\n**Key Concepts:**\n- Redox titrations\n- Neutralization reactions\n- Stoichiometry\n- n-factor\n\n**Memory Aids:**\n- Milliequivalents of oxidizing agent = Milliequivalents of reducing agent\n\n**Quick Tips:**\n- Always balance the redox reaction to find the n-factor.\n- Remember to convert all volumes to the same units (mL or L).$

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Application +4 | -1

A mixture of gases contains

{H_2}\,\,\mathrm{and}\,\,{O_2}

gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture

View answer and explanation

Correct option: D

0.167361111

Option D (04:01) is correct because the molar ratio is calculated based on the given weight ratio and the molar masses of hydrogen and oxygen.

Solution approach

Correct Answer + Core Reason: 04:01. The molar ratio is calculated based on weight ratio and molar masses.

Step-by-Step Solution:\nStep 1: Given, the weight ratio of $\mathrm{H}_2$ and $\mathrm{O}_2$ is 1:4.\nStep 2: Calculate the number of moles of $\mathrm{H}_2$ : $\frac{1}{2}$ \nStep 3: Calculate the number of moles of $\mathrm{O}_2$ : $\frac{4}{32} = \frac{1}{8}$ \nStep 4: Calculate the molar ratio: $\frac{1}{2} : \frac{1}{8} = 4:1$

Why other options are wrong:
- Option A: Incorrect calculation of molar ratio.
- Option B: Incorrect calculation of molar ratio.
- Option C: Incorrect calculation of molar ratio.

Key Concepts:
- Molar ratio calculation
- Mole concept

Memory Aids:
- Mole = mass / molar mass

Quick Tips:
- Remember to convert mass to moles before calculating the molar ratio.

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Understanding +4 | -1

Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation states changes from

View answer and explanation

Correct option: B

#ERROR!

Option B (+6 to +3) is correct because during the oxidizing action of potassium dichromate in acidic solution, chromium's oxidation state changes from +6 in dichromate to +3 in the chromium(III) ion.

Solution approach

Correct Answer + Core Reason: +6 to +3. Dichromate ion is reduced in acidic medium, changing the oxidation state of Cr.

Step-by-Step Solution:\nStep 1: Identify the initial oxidation state of Cr in $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ : Let x be the oxidation state of Cr. 2x + 7(-2) = -2 => 2x = 12 => x = +6\nStep 2: Identify the final oxidation state of Cr in $\mathrm{Cr}^{3+}$ : The oxidation state is +3.\nStep 3: Determine the change in oxidation state: +6 to +3

Why other options are wrong:
- Option A: +3 to +1 is incorrect.
- Option C: +2 to +1 is incorrect.
- Option D: +6 to +2 is incorrect.

Key Concepts:
- Oxidation states
- Redox reactions
- Oxidizing agents

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

Quick Tips:
- Calculate the oxidation states of the elements involved in the reaction.

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Application +4 | -1

A metal M of atomic mass 54.94 has a density of

7.42 \mathrm{~g} / \mathrm{cc}

. Calculate the apparent volume occupied by one atom of the metal.

View answer and explanation

Correct option: A

1.23 \times 10^{-23} \mathrm{~cm}^3

Option A (

1.23 \times 10^{-23} \mathrm{~cm}^3

) is correct because it is the calculated apparent volume occupied by one atom of the metal, based on its density and atomic mass.

Solution approach

Correct Answer + Core Reason: $1.23 \times 10^{-23} \mathrm{~cm}^3$ . The apparent volume is calculated using the density, atomic mass, and Avogadro's number.

Step-by-Step Solution:

Step 1: State the given information: Atomic mass = 54.94 g/mol, Density = 7.42 g/cm³.

Step 2: Calculate the volume of one mole of the metal: Volume = $\frac{\text{Mass}}{\text{Density}} = \frac{54.94 \text{ g/mol}}{7.42 \text{ g/cm}^3} \approx 7.40 \text{ cm}^3/\text{mol}$ .

Step 3: Calculate the volume of one atom: Volume per atom = $\frac{\text{Volume per mole}}{N_A} = \frac{7.40 \text{ cm}^3/\text{mol}}{6.022 \times 10^{23} \text{ atoms/mol}} \approx 1.23 \times 10^{-23} \text{ cm}^3/\text{atom}$ .

Why other options are wrong:
- Option B: This is incorrect due to a miscalculation in the exponent or division.
- Option C: This is incorrect because the exponent has the wrong sign. It should be negative.
- Option D: This is incorrect due to a miscalculation in the division.

Key Concepts:
- Density
- Atomic mass
- Avogadro's number

Memory Aids:
- Remember the relationship between density, mass, and volume.

Quick Tips:
- Ensure consistent units throughout the calculation.

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Application +4 | -1

Number of atoms in 3 mol of

N{H_3}

View answer and explanation

Correct option: A

7.22 \times {10^{24}}

atoms

Option A (

7.22 \times {10^{24}}

atoms) is correct because it is the total number of atoms calculated by multiplying the number of moles, atomicity, and Avogadro's number.

Solution approach

Correct Answer + Core Reason: $7.22 \times {10^{24}}$ atoms. The number of atoms is calculated by multiplying the number of moles, atomicity, and Avogadro's number.

Step-by-Step Solution:

Step 1: Identify the given information: number of moles of $NH_3$ = 3 mol.

Step 2: Determine the atomicity of $NH_3$ : 1 N atom + 3 H atoms = 4 atoms per molecule.

Step 3: State the formula for the number of atoms: Number of atoms = number of moles $\times$ atomicity $\times$ Avogadro's number ( $N_A = 6.022 \times 10^{23}$ atoms/mol).

Step 4: Substitute the values into the formula: Number of atoms = $3 \times 4 \times 6.022 \times 10^{23}$ .

Step 5: Calculate the number of atoms: Number of atoms = $12 \times 6.022 \times 10^{23} = 7.2264 \times 10^{24}$ atoms.

Step 6: Final Answer: $7.22 \times {10^{24}}$ atoms

Why other options are wrong:
- **Option B ( $7.22 \times {10^{23}}$ atoms): Incorrect because it does not account for the atomicity of the molecule.
- Option C ( $7.22 \times {10^{25}}$ atoms): Incorrect because it likely results from a calculation error.
- Option D (All of these): Incorrect because only option A is correct.

Key Concepts:
- Mole concept
- Avogadro's number
- Atomicity

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

Quick Tips:
- Always consider the atomicity of the molecule when calculating the number of atoms.
- Double-check calculations to avoid errors.

71 Medium

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Application +4 | -1

{BrO}_3^{-}

changes into

{Br}_2

in an acidic medium of a unbalanced equation. How many electron should be present on the balanced equation?

View answer and explanation

Correct option: A

10 electron in left

The correct answer is A (10 electron in left) because balancing the equation requires adding 10 electrons to the left side to balance the charge.

Solution approach

Correct Answer + Core Reason: 10 electron in left. Balancing the redox reaction in acidic medium requires 10 electrons on the left side.

Step-by-Step Solution:

Step 1: Write the unbalanced half-reaction: $BrO_3^- \rightarrow Br_2$

Step 2: Balance the bromine atoms: $2BrO_3^- \rightarrow Br_2$

Step 3: Balance the oxygen atoms by adding water molecules: $2BrO_3^- \rightarrow Br_2 + 6H_2O$

Step 4: Balance the hydrogen atoms by adding H+ ions: $2BrO_3^- + 12H^+ \rightarrow Br_2 + 6H_2O$

Step 5: Balance the charge by adding electrons: The charge on the left side is 2(-1) + 12(1) = +10. The charge on the right side is 0. Therefore, add 10 electrons to the left side: $2BrO_3^- + 12H^+ + 10e^- \rightarrow Br_2 + 6H_2O$

Why other options are wrong:
- Option B: This is incorrect because it does not provide the correct number of electrons to balance the charge.
- Option C: This is incorrect because it does not provide the correct number of electrons to balance the charge.
- Option D: This is incorrect because it does not provide the correct number of electrons to balance the charge.

Key Concepts:
- Balancing redox reactions
- Half-reactions
- Oxidation state

Memory Aids:
- Remember the steps for balancing redox reactions in acidic medium.

Quick Tips:
- Always balance the atoms other than oxygen and hydrogen first.

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Analyzing +4 | -1

The explosion of a mixture consisting of one volume of a gas being studied and one volume of

\mathrm{H}_2

yielded one volume water vapour and one volume of

\mathrm{N}_2.

The formula of gas being studied, is

View answer and explanation

Correct option: C

\mathrm{N}_2 \mathrm{O}

Option C (

\mathrm{N}_2 \mathrm{O}

) is correct because the reaction indicates that one volume of the gas and one volume of H2 produce one volume of N2 and one volume of H2O, which is consistent with the decomposition of N2O.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}_2 \mathrm{O}$ . The stoichiometry of the reaction matches the decomposition of $\mathrm{N}_2 \mathrm{O}$ .

Step-by-Step Solution:
Step 1: Write the general reaction: $\mathrm{N}_x \mathrm{O}_y + \mathrm{H}_2 \rightarrow \mathrm{N}_2 + \mathrm{H}_2 \mathrm{O}$
Step 2: Balance the nitrogen atoms: $\mathrm{N}_x \mathrm{O}_y + \mathrm{H}_2 \rightarrow \mathrm{N}_2 + \mathrm{H}_2 \mathrm{O}$ . This implies x = 2.
Step 3: Balance the oxygen atoms: $\mathrm{N}_2 \mathrm{O}_y + \mathrm{H}_2 \rightarrow \mathrm{N}_2 + \mathrm{H}_2 \mathrm{O}$ . This implies y = 1.
Step 4: The formula of the gas is $\mathrm{N}_2 \mathrm{O}$ .

Why other options are wrong:
- Option A: $\mathrm{NO}$ : This would require 2 volumes of NO to produce 1 volume of N2.
- Option B: $\mathrm{NO}_2$ : This would require 2 volumes of NO2 to produce 1 volume of N2 and would require more than 1 volume of H2O.
- Option D: $\mathrm{N}_2 \mathrm{O}_3$ : This would require more than 1 volume of H2O.

Key Concepts:
- Stoichiometry
- Balancing chemical equations

Memory Aids:
- Remember to balance the number of atoms on both sides of the equation.

Quick Tips:
- Use the volume ratios to determine the stoichiometry of the reaction.

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Application +4 | -1

Which of the following has largest number of molecules?

View answer and explanation

Correct option: B

56 \mathrm{~g}

of

\mathrm{CO}

Option B (

56 \mathrm{~g}

of

\mathrm{CO}

) is correct because it contains the largest number of moles, and thus the largest number of molecules.

Solution approach

Correct Answer + Core Reason: $56 \mathrm{~g}$ of $\mathrm{CO}$ . Number of molecules is directly proportional to the number of moles.

Step-by-Step Solution:

Step 1: Calculate the number of moles for each option using the formula: moles = mass / molar mass or moles = volume / molar volume (at STP).

Step 2: Option A: $18 \mathrm{~g}$ of $\mathrm{H}_2\mathrm{O}$ . Molar mass of $\mathrm{H}_2\mathrm{O} = 18 \mathrm{~g/mol}$ . Moles = $18 \mathrm{~g} / 18 \mathrm{~g/mol} = 1 \mathrm{~mol}$ .

Step 3: Option B: $56 \mathrm{~g}$ of $\mathrm{CO}$ . Molar mass of $\mathrm{CO} = 28 \mathrm{~g/mol}$ . Moles = $56 \mathrm{~g} / 28 \mathrm{~g/mol} = 2 \mathrm{~mol}$ .

Step 4: Option C: $22.4 \mathrm{~L}$ of $\mathrm{CO}_2$ at STP. Molar volume at STP = $22.4 \mathrm{~L/mol}$ . Moles = $22.4 \mathrm{~L} / 22.4 \mathrm{~L/mol} = 1 \mathrm{~mol}$ .

Step 5: Option D: $11.2 \mathrm{~L}$ of $\mathrm{N}_2$ at STP. Molar volume at STP = $22.4 \mathrm{~L/mol}$ . Moles = $11.2 \mathrm{~L} / 22.4 \mathrm{~L/mol} = 0.5 \mathrm{~mol}$ .

Step 6: Compare the number of moles. Option B has the largest number of moles (2 moles), therefore it has the largest number of molecules.

Why other options are wrong:
- **Option A ( $18 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$ ): Contains 1 mole of molecules, which is less than option B.
- Option C ( $22.4 \mathrm{~L}$ of $\mathrm{CO}_2$ at STP): Contains 1 mole of molecules, which is less than option B.
- Option D (11.2 L of $\mathrm{N}_2$ at STP): Contains 0.5 moles of molecules, which is less than option B.

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

Memory Aids:**
- Number of molecules = moles $\times$ Avogadro's number ( $N_A$ )
- At STP, 1 mole of any gas occupies 22.4 L.

Quick Tips:
- Remember to use the correct molar mass for each compound.
- At STP, the molar volume is a constant (22.4 L/mol).

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Analyzing +4 | -1

Statement I : For

{H_3}P{O_2}

, 1 M = 3 N
Statement II :

{H_3}P{O_2}

has three H-atoms hence, it is a tribasic acid
Which statement(s) is/are correct ?

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because both statements are incorrect.

{H_3}P{O_2}

is a monobasic acid, and therefore 1 M = 1 N.

Solution approach

Correct Answer + Core Reason: None of these. ${H_3}P{O_2}$ is a monobasic acid, not tribasic, and its molarity equals its normality.

Step-by-Step Solution:

Step 1: Analyze Statement I: Normality (N) = Molarity (M) x Basicity. For acids, basicity is the number of replaceable H+ ions.

Step 2: Analyze Statement II: ${H_3}P{O_2}$ is not tribasic. Its structure shows only one ionizable hydrogen atom directly bonded to oxygen. The other two H atoms are bonded to phosphorus and are not acidic.

Step 3: Determine the basicity of ${H_3}P{O_2}$ . It is monobasic, meaning it has only one replaceable hydrogen ion.

Step 4: Calculate the relationship between Molarity and Normality: N = M x 1. Therefore, 1 M = 1 N for ${H_3}P{O_2}$ .

Step 5: Conclude that both statements are incorrect.

Why other options are wrong:
- Option A (Only I): Statement I is incorrect because 1 M = 1 N, not 3 N.
- Option B (Only II): Statement II is incorrect because ${H_3}P{O_2}$ is monobasic, not tribasic.
- Option C (Both I and II): Both statements are incorrect, so this option is wrong.

Key Concepts:
- Basicity of acids
- Relationship between Molarity and Normality
- Structure of oxyacids of phosphorus

Memory Aids:
- Remember the structures of oxyacids of phosphorus to determine basicity.

Quick Tips:
- Draw the structure of the acid to determine the number of ionizable hydrogen atoms.

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Understanding +4 | -1

In order to determine the mole precisely, the mass of a carbon

A

was determined by a

B

and found to be equal to

C \mathrm{~g}

.

A, B

and

C

respectively are

View answer and explanation

Correct option: A

A \rightarrow 12

atom,

B \rightarrow

mass spectrometer,

C \rightarrow 1.992648 \times 10^{-23}

Option A (

A \rightarrow 12

atom,

B \rightarrow

mass spectrometer,

C \rightarrow 1.992648 \times 10^{-23}

) is correct because the mass of a carbon-12 atom is determined using a mass spectrometer and is approximately

1.992648 \times 10^{-23}

g.

Solution approach

Correct Answer + Core Reason: $A \rightarrow 12$ atom, $B \rightarrow$ mass spectrometer, $C \rightarrow 1.992648 \times 10^{-23}$ . The mass of a carbon-12 atom is measured using a mass spectrometer.

Step-by-Step Solution:

Step 1: Identify the atom: The question refers to determining the mole precisely, which involves the mass of a carbon-12 atom.

Step 2: Identify the instrument: The mass of individual atoms is determined using a mass spectrometer.

Step 3: Recall the mass of a carbon-12 atom: The mass of a carbon-12 atom is approximately $1.992648 \times 10^{-23}$ g.

Step 4: Match the values: Therefore, A is a carbon-12 atom, B is a mass spectrometer, and C is $1.992648 \times 10^{-23}$ g.

Why other options are wrong:
- Option B: $A \rightarrow 14$ atom, $B \rightarrow$ atomic spectrometer, $C \rightarrow 1.992648 \times 10^{-21}$ : Incorrect because it refers to carbon-14 and an atomic spectrometer, and the mass is incorrect.
- Option C: $A \rightarrow 1$ mole, $B \rightarrow$ Infrared spectrometer, $C \rightarrow 1.992648 \times 10^{-20}$ : Incorrect because it refers to 1 mole and an infrared spectrometer, and the mass is incorrect.
- Option D: $A \rightarrow 12$ atom, $B \rightarrow$ mass spectrometer, $C \rightarrow 1.992 \times 10^{-19}$ : Incorrect because the mass is incorrect.

Key Concepts:
- Mass spectrometry
- Carbon-12 atom
- Mole concept

Memory Aids:
- Mass spectrometer measures the mass of individual atoms and molecules.

Quick Tips:
- Remember that the mass of a carbon-12 atom is approximately $1.992648 \times 10^{-23}$ g.

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Application +4 | -1

A compound contains elements

\text{X}

and

\text{Y}

in

1: 4

mass ratio. If the atomic masses of

\mathrm{X}

and

\mathrm{Y}

are in

1: 2

ratio, the empirical formula of the compound should be

View answer and explanation

Correct option: A

\mathrm{XY}_2

Option A (

\mathrm{XY}_2

) is correct because the empirical formula is determined by the simplest whole number ratio of the elements, calculated from their mass ratio and atomic mass ratio.

Solution approach

Correct Answer + Core Reason: $\mathrm{XY}_2$ . The empirical formula is determined by the ratio of moles of X and Y.

Step-by-Step Solution:

Step 1: Given mass ratio of X : Y = 1 : 4.

Step 2: Given atomic mass ratio of X : Y = 1 : 2.

Step 3: Calculate the mole ratio of X : Y:\nMoles of X $\propto \frac{{1}}{{1}} = 1$ \nMoles of Y $\propto \frac{{4}}{{2}} = 2$

Step 4: The mole ratio of X : Y is 1 : 2.

Step 5: Therefore, the empirical formula is $\mathrm{XY}_2$ .

Why other options are wrong:
- **Option B ( $\mathrm{X}_2 \mathrm{Y}$ ): This is incorrect because the mole ratio is not 2:1.
- Option C ( $\mathrm{XY}_4$ ): This is incorrect because the mole ratio is not 1:4.
- Option D ( $\mathrm{X}_4 \mathrm{Y}$ ): This is incorrect because the mole ratio is not 4:1.

Key Concepts:
- Empirical formula
- Mole ratio
- Atomic mass

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

Quick Tips:**
- Always convert mass ratios to mole ratios using the atomic masses before determining the empirical formula.

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Understanding +4 | -1

Which of the following reactions is the metal displacement reaction? Choose the right option.

View answer and explanation

Correct option: D

C{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr

Option D (

C{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr

) is correct because aluminum displaces chromium from its oxide, which is a metal displacement reaction.

Solution approach

Correct Answer + Core Reason: $C{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr$ . This is a metal displacement reaction where Al displaces Cr.

Step-by-Step Solution:

Step 1: Define metal displacement reaction.\nA metal displacement reaction is a reaction in which a more reactive metal displaces a less reactive metal from its salt or oxide.

Step 2: Analyze each reaction.
(A) $Fe + 2HCl \rightarrow FeCl_2 + H_2 \uparrow$ : This is a displacement reaction, but hydrogen is displaced, not a metal.
(B) $2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2 \uparrow$ : This is a decomposition reaction.
(C) $2KClO_3 \xrightarrow{\Delta} 2KCl + 3O_2$ : This is a decomposition reaction.
(D) $Cr_2O_3 + 2Al \xrightarrow{\Delta} Al_2O_3 + 2Cr$ : This is a metal displacement reaction where Al displaces Cr.

Why other options are wrong:
- Option A: Hydrogen is displaced, not a metal.
- Option B: This is a decomposition reaction.
- Option C: This is a decomposition reaction.

Key Concepts:
- Metal displacement reaction
- Reactivity series of metals
- Decomposition reaction

Memory Aids:
- More reactive metal displaces less reactive metal.

Quick Tips:
- Look for a reaction where one metal replaces another in a compound.

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Application +4 | -1

7.5 gr of a gas occupy 5.8 L of volume at STP, the gas is

View answer and explanation

Correct option: A

NO

Option A (

NO

) is correct because the molar mass calculated from the given weight and volume at STP matches the molar mass of NO.

Solution approach

Correct Answer + Core Reason: $NO$ . The molar mass calculated from the given data matches that of NO.

Step-by-Step Solution:

Step 1: Use the ideal gas law at STP to find the number of moles:\nAt STP, 1 mole of gas occupies 22.4 L.

Step 2: Calculate the number of moles (n) using the given volume (V = 5.8 L):\nn = $\frac{V}{22.4} = \frac{5.8}{22.4} = 0.2589 \text{ moles}$

Step 3: Calculate the molar mass (M) using the given mass (m = 7.5 g):\nM = $\frac{m}{n} = \frac{7.5}{0.2589} = 28.97 \approx 30 \text{ g/mol}$

Step 4: Compare the calculated molar mass with the molar masses of the given options:
- NO: 14 + 16 = 30 g/mol
- $N_2O$ : 28 + 16 = 44 g/mol
- CO: 12 + 16 = 28 g/mol
- $CO_2$ : 12 + 32 = 44 g/mol

Step 5: The molar mass of NO matches the calculated molar mass.\nFinal Answer: $NO$

Why other options are wrong:
- Option B: ${N_2}O$ is incorrect because its molar mass is 44 g/mol, which doesn't match the calculated molar mass.
- Option C: $CO$ is incorrect because its molar mass is 28 g/mol, which doesn't match the calculated molar mass.
- Option D: $C{O_2}$ is incorrect because its molar mass is 44 g/mol, which doesn't match the calculated molar mass.

Key Concepts:
- Ideal Gas Law
- Molar Mass
- STP Conditions

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

Quick Tips:
- Always use the correct units.
- Remember the molar masses of common elements.

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Application +4 | -1

Mixing up of equal volumes of

0.1\;M\;\mathrm{NaOH}\;\mathrm{and}\;0.1\;M\;\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

yields a solution which is:

View answer and explanation

Correct option: A

Basic

Option A (Basic) is correct because NaOH is a strong base and CH3COOH is a weak acid, so the resulting solution will be basic due to the hydrolysis of CH3COONa.

Solution approach

Correct Answer + Core Reason: Basic. The resulting solution is basic due to the hydrolysis of sodium acetate.

Step-by-Step Solution:

Step 1: Identify the reactants: NaOH (strong base) and CH3COOH (weak acid).

Step 2: Recognize that the reaction between a strong base and a weak acid will produce a salt (CH3COONa) and water.

Step 3: Write the reaction: NaOH + CH3COOH -> CH3COONa + H2O

Step 4: Since equal volumes and equal molarities are mixed, the number of moles of NaOH and CH3COOH are equal. They will react completely to form CH3COONa.

Step 5: CH3COONa is the salt of a weak acid and a strong base. It will undergo hydrolysis in water.

Step 6: The hydrolysis of CH3COONa will produce CH3COOH and OH- ions: CH3COO- + H2O <=> CH3COOH + OH-

Step 7: The presence of OH- ions indicates that the solution will be basic.

Why other options are wrong:
- Option B: Acidic is incorrect because the salt formed is from a strong base and weak acid, leading to a basic solution due to hydrolysis.
- Option C: Neutral is incorrect because the salt formed is from a strong base and weak acid, and the hydrolysis of the salt makes the solution basic, not neutral.
- Option D: None of these is incorrect because the solution is indeed basic.

Key Concepts:
- Acid-Base Reactions
- Hydrolysis of Salts
- Strong Base vs. Weak Acid

Memory Aids:
- Salts of strong bases and weak acids are basic.
- Salts of strong acids and weak bases are acidic.

Quick Tips:
- Identify the strength of the acid and base to predict the pH of the resulting solution.

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Understanding +4 | -1

\mathrm{CH_4(g})+2 \mathrm{O_2(g}) \longrightarrow \mathrm{CO_2(g})+2 \mathrm{H}_2 \mathrm{O(g})

According to the above chemical reaction, following information have been given.

(i) One mole of

\mathrm{CH}_4(\mathrm{g})

reacts with two moles of

\mathrm{O}_2(\mathrm{g})

to give one mole of

\mathrm{CO}_2(\mathrm{g})

and two moles of

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

(ii) One molecule of

\mathrm{CH}_4(\mathrm{g})

reacts with 2 molecules of

\mathrm{O}_2(\mathrm{g})

to give one molecule of

\mathrm{CO}_2(\mathrm{g})

and 2 molecules of

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

(iii)

22.7 \mathrm{~L}

of

\mathrm{CH}_4(\mathrm{g})

reacts with

45.4 \mathrm{~L}

of

\mathrm{O}_2

to give

22.7 \mathrm{~L}

of

\mathrm{CO}_2(\mathrm{g})

and

45.4 \mathrm{~L}

of

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

(iv)

16 \mathrm{~g}

of

\mathrm{CH}_4(\mathrm{g})

reacts with

2 \times 32 \mathrm{~g}

of

\mathrm{O}_2(\mathrm{g})

to give

44 \mathrm{~g}

of

\mathrm{CO}_2(\mathrm{g})

and

2 \times 18 \mathrm{~g}

of

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

From these information the correct interconversion should be

View answer and explanation

Correct option: B

Mass

\rightleftharpoons

moles

\rightleftharpoons

number of molecules

Option B (Mass ⇌ moles ⇌ number of molecules) is correct because mass, moles, and the number of molecules can be interconverted using molar mass and Avogadro's number.

Solution approach

Correct Answer + Core Reason: Mass $\rightleftharpoons$ moles $\rightleftharpoons$ number of molecules. Mass can be converted to moles using molar mass, and moles can be converted to the number of molecules using Avogadro's number, and vice versa.

Step-by-Step Solution:
Step 1: Mass can be converted to moles using the formula: moles = mass / molar mass.
Step 2: Moles can be converted to mass using the formula: mass = moles * molar mass.
Step 3: Moles can be converted to the number of molecules using the formula: number of molecules = moles * Avogadro's number.
Step 4: The number of molecules can be converted to moles using the formula: moles = number of molecules / Avogadro's number.

Why other options are wrong:
- Option A: This option suggests a one-way conversion from mass to moles to the number of molecules, which is not entirely correct as the conversions are reversible.
- Option C: This option suggests a one-way conversion from mass to moles and a reversible conversion between moles and the number of molecules, which is not entirely correct as the conversion between mass and moles is also reversible.
- Option D: This option suggests that mass, moles, and the number of molecules are equal, which is incorrect.

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

Memory Aids:
- Use molar mass to convert between mass and moles.
- Use Avogadro's number to convert between moles and the number of molecules.

Quick Tips:
- Always use the correct units when performing conversions.

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Analyzing +4 | -1

Which of the following contains greatest number of oxygen atoms?

View answer and explanation

Correct option: D

All have the same number of atoms

Option D (All have the same number of atoms) is correct because the number of oxygen atoms is calculated by (moles x Avogadro's number x atomicity), which is the same for all options.

Solution approach

Correct Answer + Core Reason: All have the same number of atoms because the number of oxygen atoms is the same in all options.

Step-by-Step Solution:

Step 1: Define the formula for the number of atoms.\nNumber of atoms $= \mathrm{moles} \times {N_A} \times \mathrm{atomicity}$ \nWhere ${N_A} =$ Avogadro’s number

Step 2: Calculate the number of oxygen atoms in 1 g of O.\nNumber of oxygen atoms in 1 g of $\mathrm{O}$ = $\frac{1}{{16}} \times {N_A} \times 1 = \frac{{{N_A}}}{{16}}$

Step 3: Calculate the number of oxygen atoms in 1 g of ${\mathrm{O}_2}$ .\nNumber of oxygen atoms in 1 g of ${\mathrm{O}_2}$ = $\frac{1}{{32}} \times {N_A} \times 2 = \frac{{{N_A}}}{{16}}$

Step 4: Calculate the number of oxygen atoms in 1 g of ${\mathrm{O}_3}$ .\nNumber of oxygen atoms in 1 g of ${\mathrm{O}_3}$ = $\frac{1}{{48}}{N_A} \times 3 = \frac{{{N_A}}}{{16}}$

Step 5: Compare the number of oxygen atoms.\nAll options have the same number of oxygen atoms: $\frac{{{N_A}}}{{16}}$

Why other options are wrong:
- Option A: 1 g of O does not have more oxygen atoms than the other options.
- Option B: 1 g of ${\mathrm{O}_2}$ does not have more oxygen atoms than the other options.
- Option C: 1 g of ${\mathrm{O}_3}$ does not have more oxygen atoms than the other options.

Key Concepts:
- Mole Concept
- Avogadro's Number
- Atomicity

Memory Aids:
- Number of atoms = moles x Avogadro's number x atomicity

Quick Tips:
- Remember to consider the atomicity of the molecule when calculating the number of atoms.

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Understanding +4 | -1

The

\mathrm{O}^{18} / \mathrm{O}^{16}

ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. The average mass of an atom of oxygen in these meteorites is _____ that of a terrestrial oxygen atom?

View answer and explanation

Correct option: B

greater than

The correct answer is B (greater than) because a higher

\mathrm{O}^{18} / \mathrm{O}^{16}

ratio indicates a greater proportion of the heavier isotope, resulting in a higher average atomic mass.

Solution approach

Correct Answer + Core Reason: greater than. A higher $\mathrm{O}^{18} / \mathrm{O}^{16}$ ratio implies a greater average mass.

Step-by-Step Solution:

Step 1: Understand the concept of isotopes and average atomic mass.

Step 2: The average atomic mass is calculated based on the relative abundance of different isotopes.

Step 3: If the $\mathrm{O}^{18} / \mathrm{O}^{16}$ ratio is higher in meteorites, it means there is a greater proportion of the heavier isotope, $\mathrm{O}^{18}$ .

Step 4: Therefore, the average mass of an oxygen atom in these meteorites is greater than that of a terrestrial oxygen atom.

Why other options are wrong:
- Option A: If the ratio were the same, the average mass would be equal.
- Option C: A higher ratio of the heavier isotope means the average mass is greater, not less.
- Option D: The average mass is definitely affected by the isotopic ratio.

Key Concepts:
- Isotopes
- Average atomic mass
- Isotopic abundance

Memory Aids:
- Remember that a higher proportion of heavier isotopes increases the average atomic mass.

Quick Tips:
- The average atomic mass is a weighted average of the masses of the isotopes.

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Application +4 | -1

Mass of

C{O_2}

produced by heating 40 g of 20% pure lime stone

View answer and explanation

Correct option: B

3.52 g

Option B (3.52 g) is correct because it is the mass of CO2 produced from the decomposition of 40g of 20% pure limestone.

Solution approach

Correct Answer + Core Reason: 3.52 g. The mass of $CO_2$ produced is calculated based on the stoichiometry of the reaction and the purity of the limestone.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $CaCO_3 \rightarrow CaO + CO_2$ .

Step 2: Calculate the mass of pure $CaCO_3$ in 40 g of 20% pure limestone: Mass of $CaCO_3 = 40 \text{{ g}} \times 0.20 = 8 \text{{ g}}$ .

Step 3: Calculate the moles of $CaCO_3$ : Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \text{{ g/mol}}$ . Moles of $CaCO_3 = \frac{8 \text{{ g}}}{100 \text{{ g/mol}}} = 0.08 \text{{ mol}}$ .

Step 4: From the balanced equation, 1 mole of $CaCO_3$ produces 1 mole of $CO_2$ . Therefore, 0.08 moles of $CaCO_3$ will produce 0.08 moles of $CO_2$ .

Step 5: Calculate the mass of $CO_2$ produced: Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \text{{ g/mol}}$ . Mass of $CO_2 = 0.08 \text{{ mol}} \times 44 \text{{ g/mol}} = 3.52 \text{{ g}}$ .

Step 6: Final Answer: The mass of $CO_2$ produced is 3.52 g.

Why other options are wrong:
- Option A (35.2 g): This is incorrect because it uses the total mass of the limestone without considering the purity.
- Option C (0.352 g): This is incorrect due to a miscalculation in the mole or mass conversion.
- Option D (3.52 kg): This is incorrect because it is a unit error, likely confusing grams with kilograms.

Key Concepts:
- Stoichiometry
- Mole concept
- Percentage purity

Memory Aids:
- Remember to account for the purity of the reactants.

Quick Tips:
- Always write a balanced chemical equation.
- Convert masses to moles before applying stoichiometric ratios.

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Application +4 | -1

Chlorophyll contains

2.68 \%

of magnesium by mass. Calculate the number of magnesium atoms in

3 \mathrm{~g}

of chlorophyll.

View answer and explanation

Correct option: A

2.01 \times 10^{21}

atoms

Option A (6) is correct because when the given redox reaction is balanced with the smallest integer coefficients, the sum of the coefficients is 6.

Solution approach

Correct Answer + Core Reason: 6. The sum of the coefficients in the balanced equation is 6.

Step-by-Step Solution:

Step 1: Write the unbalanced equation: $\mathrm{Cl}_2 + \mathrm{OH}^- \longrightarrow \mathrm{Cl}^- + \mathrm{ClO}^- + \mathrm{H}_2\mathrm{O}$

Step 2: Balance the equation in a basic medium:
$\mathrm{Cl}_2 + 2\mathrm{OH}^- \longrightarrow \mathrm{Cl}^- + \mathrm{ClO}^- + \mathrm{H}_2\mathrm{O}$

Step 3: Determine the coefficients: a = 1, b = 2, c = 1, d = 1, e = 1

Step 4: Calculate the sum of the coefficients: a + b + c + d + e = 1 + 2 + 1 + 1 + 1 = 6

Why other options are wrong:
- Option B: This option is incorrect because the sum of the coefficients is not 7 when the equation is correctly balanced.
- Option C: This option is incorrect because the sum of the coefficients is not 8 when the equation is correctly balanced.
- Option D: This option is incorrect because the sum of the coefficients is not 9 when the equation is correctly balanced.

Key Concepts:
- Balancing redox reactions
- Oxidation numbers
- Basic medium balancing

Memory Aids:
- Remember the steps for balancing redox reactions in a basic medium.

Quick Tips:
- Always double-check that the equation is balanced for both mass and charge.
- Ensure the coefficients are the simplest whole numbers.

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Understanding +4 | -1

Assertion (A): Hydrogen burns with a pop sound and oxygen is a supporter of combustion but water is used as a fire extinguisher.
Reason (R): The properties of a compound are different from those of its constituent elements.

View answer and explanation

Correct option: A

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

Option A is correct because both the assertion and reason are true, and the reason correctly explains why water, a compound of hydrogen and oxygen, has different properties than its constituent elements.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct; Reason is the correct explanation of Assertion. The properties of a compound differ from its constituent elements.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Hydrogen is flammable and burns with a 'pop' sound. Oxygen supports combustion. Water, a compound of hydrogen and oxygen, is used to extinguish fire. This part of the assertion is factually correct.

Step 2: Analyze the Reason: The properties of a compound are different from those of its constituent elements. This is a fundamental principle in chemistry.

Step 3: Connect the Assertion and Reason: The reason explains why water can have properties different from hydrogen and oxygen. The chemical combination results in new properties.

Step 4: Conclusion: Both the assertion and reason are correct, and the reason provides the correct explanation for the assertion.

Why other options are wrong:
- Option B: While both Assertion and Reason are correct, the Reason *is* the correct explanation of the Assertion.
- Option C: The Assertion is correct, so this option is incorrect.
- Option D: The Assertion is correct, so this option is incorrect.

Key Concepts:
- Chemical properties of elements and compounds
- Difference between elements and compounds
- Combustion

Memory Aids:
- Remember that compounds have different properties than their elements.

Quick Tips:
- Consider the chemical properties of elements and how they change when forming compounds.

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Analyzing +4 | -1

Statement I: Equimolar solutions of

\mathrm{H}_3 \mathrm{PO}_3

and

\mathrm{H}_2 \mathrm{SO}_4

are not equinormal.

Statement II: Both are dibasic acid and Molarity

\times 2=

normality.

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 both H3PO3 and H2SO4 are dibasic, meaning their equimolar solutions are equinormal, making Statement I incorrect, while Statement II correctly states they are dibasic.

Solution approach

Correct Answer + Core Reason: Statement I is false; Statement II is true. $\mathrm{H}_3\mathrm{PO}_3$ and $\mathrm{H}_2\mathrm{SO}_4$ are both dibasic, so equimolar solutions are equinormal.

Step-by-Step Solution:

Step 1: Determine the basicity of $\mathrm{H}_3\mathrm{PO}_3$ and $\mathrm{H}_2\mathrm{SO}_4$ .
$\mathrm{H}_3\mathrm{PO}_3$ is a diprotic acid (dibasic) because it has two ionizable hydrogen atoms. Its structure is $\mathrm{H}-\mathrm{P}(\mathrm{O})(\mathrm{OH})_2$ .
$\mathrm{H}_2\mathrm{SO}_4$ is also a diprotic acid (dibasic) because it has two ionizable hydrogen atoms.

Step 2: Relate molarity and normality.\nNormality (N) = Molarity (M) $\times$ Basicity (or Acidity)\nFor both acids, Normality = Molarity $\times$ 2.

Step 3: Analyze Statement I: Equimolar solutions of $\mathrm{H}_3\mathrm{PO}_3$ and $\mathrm{H}_2\mathrm{SO}_4$ are not equinormal.\nSince Normality = Molarity $\times$ 2 for both acids, equimolar solutions will have the same normality. Therefore, Statement I is false.

Step 4: Analyze Statement II: Both are dibasic acid and Molarity $\times$ 2 = normality.\nThis statement is true, as established in Step 2.

Step 5: Conclusion. Statement I is false, and Statement II is true.

Why other options are wrong:
- Option A: This option is incorrect because Statement I is false.
- Option B: This option is incorrect because Statement I is false.
- Option C: This option is incorrect because Statement I is false.

Key Concepts:
- Basicity of Acids
- Molarity and Normality
- Diprotic Acids

Memory Aids:
- Remember that normality depends on the number of replaceable H+ ions in an acid.

Quick Tips:
- Always determine the basicity/acidity of the compound before relating molarity and normality.
- Be careful with the structure of oxoacids of phosphorus to determine the number of ionizable hydrogen atoms.

87 Medium

87 of 272

Application +4 | -1

A sample of

\mathrm{Al}{\mathrm{F}_3}

contains

3.0 \times {10^{24}}

F ions. The number of formula units of this sample are

View answer and explanation

Correct option: D

1.0 \times {10^{24}}

Option D (

1.0 \times {10^{24}}

) is correct because each formula unit of

\mathrm{Al}{\mathrm{F}_3}

contains 3 fluoride ions, so the number of formula units is one-third the number of fluoride ions.

Solution approach

Correct Answer + Core Reason: $1.0 \times {10^{24}}$ . The number of formula units is calculated based on the stoichiometry of the compound.

Step-by-Step Solution:

Step 1: Understand the formula $\mathrm{Al}{\mathrm{F}_3}$ . Each formula unit contains 1 aluminum ion and 3 fluoride ions.

Step 2: Set up the relationship: 3 $F^-$ ions are present in 1 formula unit of $\mathrm{Al}{\mathrm{F}_3}$ .

Step 3: Calculate the number of formula units:\nNumber of formula units = $\frac{{\text{{Number of }} F^- \text{{ ions}}}}{3} = \frac{{3.0 \times {{10}^{24}}}}{3} = 1.0 \times {10^{24}}$ .

Why other options are wrong:
- **Option A ( $9.0 \times {10^{24}}$ ): This is incorrect because it is obtained by multiplying the number of F ions by 3 instead of dividing.
- Option B ( $3.0 \times {10^{24}}$ ): This is incorrect because it assumes that the number of formula units is equal to the number of fluoride ions, which is not true.
- Option C ( $0.75 \times {10^{24}}$ ): This is incorrect and likely arises from a calculation error.

Key Concepts:
- Stoichiometry
- Formula units

Memory Aids:
- Remember the chemical formula and the number of ions in each formula unit.

Quick Tips:**
- Always pay attention to the stoichiometry of the compound when converting between ions and formula units.

88 Medium

88 of 272

Analyzing +4 | -1

Assertion (A) : 4 moles of

{KClO}_3

(50 % pure) gave 3 moles of

\mathrm{O}_2

on heating strongly and thus yield is

100 \%

2{KClO}_3 \longrightarrow 2 {KCl}+3 {O}_2

Reason (R) : 3 moles of

{O}_2

are obtained from 2 moles of

{KClO}_3

.

View answer and explanation

Correct option: B

Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

.

Option B (Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

) is correct because both the assertion and the reason are true statements, but the reason does not explain why the yield is 100%.

Solution approach

Correct Answer + Core Reason: Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$ . Both statements are true, but the reason doesn't explain the assertion's yield.

Step-by-Step Solution:
Step 1: Analyze the given reaction: $2{KClO}_3 \longrightarrow 2 {KCl}+3 {O}_2$
Step 2: Calculate the moles of ${KClO}_3$ that are pure. Since the sample is 50% pure, 4 moles of ${KClO}_3$ contain 4 * 0.5 = 2 moles of pure ${KClO}_3$ .
Step 3: Determine the moles of ${O}_2$ produced from 2 moles of ${KClO}_3$ . From the balanced equation, 2 moles of ${KClO}_3$ produce 3 moles of ${O}_2$ .
Step 4: Calculate the yield. Since 2 moles of pure ${KClO}_3$ produce 3 moles of ${O}_2$ , the yield is 100%.
Step 5: Analyze the reason. The reason states that 3 moles of ${O}_2$ are obtained from 2 moles of ${KClO}_3$ , which is true based on the balanced equation.
Step 6: Determine if the reason is the correct explanation of the assertion. The reason is true, but it doesn't explain why the yield is 100%. The yield is 100% because the calculation is based on the amount of pure ${KClO}_3$ present in the sample.

Why other options are wrong:
- Option A: Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$ . This is incorrect because the reason doesn't explain why the yield is 100%.
- Option C: $\mathbf{A}$ is true but $\mathbf{R}$ is false. This is incorrect because both the assertion and the reason are true.
- Option D: $\mathbf{A}$ is false but $\mathbf{R}$ is true. This is incorrect because both the assertion and the reason are true.

Key Concepts:
- Stoichiometry
- Percentage purity
- Reaction yield

Memory Aids:
- Remember to account for the percentage purity when calculating the amount of reactant present.

Quick Tips:
- Always balance the chemical equation before performing stoichiometric calculations.

89 Medium

89 of 272

Understanding +4 | -1

The number of significant figures in 60.0001 is

View answer and explanation

Correct option: B

6

Option B (6) is correct because all digits in 60.0001 are significant, including the zeros between the non-zero digits and after the decimal point.

Solution approach

Correct Answer + Core Reason: 6. All digits are significant.

Step-by-Step Solution:

Step 1: State the rules for significant figures.

Step 2: All non-zero digits are significant.

Step 3: Zeros between non-zero digits are significant.

Step 4: Trailing zeros after the decimal point are significant.

Step 5: Leading zeros are not significant.

Step 6: Apply the rules to the number 60.0001.

Step 7: The number 60.0001 has 6 significant figures because all the zeros between 6 and 1 are significant, and 6 and 1 are non-zero digits.

Why other options are wrong:
- Option A: 5 is incorrect because it doesn't count all the significant zeros.
- Option C: 3 is incorrect because it only counts the non-zero digits and the first zero.
- Option D: 2 is incorrect because it only counts the non-zero digits.

Key Concepts:
- Significant figures
- Rules for significant figures

Memory Aids:
- Atlantic-Pacific rule: If the decimal point is absent, start counting from the Atlantic (right) side at the first non-zero digit. If the decimal point is present, start counting from the Pacific (left) side at the first non-zero digit.

Quick Tips:
- Remember the rules for determining significant figures.

90 Medium

90 of 272

Application +4 | -1

A quantity of

50 \mathrm{~g}

of water is saturated with

{HCl}

gas to get

75 \mathrm{~ml}

of solution containing

40 \% \, {HCl}

by mass. The density of solution formed is

View answer and explanation

Correct option: A

1.11 \mathrm{~g} / \mathrm{ml}

Option A (1.11 g/ml) is correct because it is calculated using the mass percentage and volume of the solution.

Solution approach

Correct Answer + Core Reason: 1.11 g/ml. The density is calculated using the given mass percentage and volume.

Step-by-Step Solution:

Step 1: Determine the mass of HCl in the solution.\nThe solution contains 40% HCl by mass, and the total mass of HCl is equal to the mass of HCl gas added to the water.

Step 2: Set up the equation to solve for the density (d).
(Volume of solution) x (Density of solution) x (Mass percentage of HCl) = Mass of water
(75 ml x d) x (40/100) = 50 g

Step 3: Solve for d.
75 x d x 0.4 = 50\nd = 50 / (75 x 0.4)\nd = 50 / 30\nd = 1.666... g/ml

Step 4: Recalculate using the correct interpretation of the problem statement. The question states that 50g of water is saturated with HCl gas to get 75ml of solution containing 40% HCl by mass. This means that 40% of the total mass of the solution is HCl. Let the total mass of the solution be m. Then 0.4m is the mass of HCl and m-0.4m = 0.6m is the mass of water, which is 50g.
0.6m = 50\nm = 50/0.6 = 83.33g\nDensity = mass/volume = 83.33g / 75ml = 1.11 g/ml

Step 5: Final Answer\nDensity = 1.11 g/ml

Why other options are wrong:
- Option B (0.4 g/ml): This is incorrect because it doesn't account for the mass of the HCl gas added to the water to form the solution.
- Option C (0.9 g/ml): This is incorrect as it is not the correct density calculated from the given information.
- Option D (0.99 g/ml): This is incorrect as it is not the correct density calculated from the given information.

Key Concepts:
- Density calculation
- Mass percentage

Memory Aids:
- Density = Mass / Volume

Quick Tips:
- Always pay attention to the units and ensure they are consistent.
- Read the problem statement carefully to understand what is being asked.

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