SOME BASIC CONCEPTS OF CHEMISTRY MCQs for JEE CHEMISTRY — Medium, Page 8

211 Medium

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Application +4 | -1

Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is

View answer and explanation

Correct option: D

6

Option D (6) is correct because 6 moles of Mohr's salt are required to reduce 1 mole of dichromate, based on the balanced redox reaction.

Solution approach

Correct Answer + Core Reason: 6. The number of moles of Mohr's salt required is determined by the stoichiometry of the redox reaction.

Step-by-Step Solution:

Step 1: Write the balanced redox reaction between potassium dichromate and Mohr's salt:
$6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$

Step 2: Identify the oxidation states of the elements involved. Chromium changes from +6 in $Cr_2O_7^{2-}$ to +3 in $Cr^{3+}$ . Iron changes from +2 in $Fe^{2+}$ to +3 in $Fe^{3+}$ .

Step 3: Determine the number of electrons transferred. Each chromium atom gains 3 electrons (total 6 electrons for $Cr_2O_7^{2-}$ ), and each iron atom loses 1 electron.

Step 4: From the balanced equation, it is clear that 6 moles of $Fe^{2+}$ (Mohr's salt) are required to reduce 1 mole of $Cr_2O_7^{2-}$ (dichromate).

Step 5: Therefore, the number of moles of Mohr's salt required per mole of dichromate is 6.

Why other options are wrong:
- Option A (3): This is incorrect. It might be a result of considering only one chromium atom gaining 3 electrons, but dichromate has two chromium atoms.
- Option B (4): This is incorrect. It might be a result of some incorrect balancing or misunderstanding of the electron transfer.
- Option C (5): This is incorrect. It might be a result of some incorrect balancing or misunderstanding of the electron transfer.

Key Concepts:
- Redox titration
- Balancing redox reactions
- Oxidation states
- Stoichiometry

Memory Aids:
- Remember the oxidation states of common ions.
- OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).

Quick Tips:
- Always balance the redox reaction before determining the mole ratio.
- Identify the oxidation states of the elements involved in the reaction.

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Application +4 | -1

Calculate the mass of lime

(\mathrm{CaO})

obtained by heating

400 \mathrm{~g}

of

80 \%

pure lime stone

\left(\mathrm{CaCO}_3\right)

View answer and explanation

Correct option: A

179.2 \mathrm{~g}

Option A (

179.2 \mathrm{~g}

) is correct because it is calculated by first finding the mass of pure

CaCO_3

and then using stoichiometry to determine the mass of CaO produced.

Solution approach

Correct Answer + Core Reason: $179.2 \mathrm{~g}$ . The mass of lime is calculated using stoichiometry based on the amount of pure calcium carbonate.

Step-by-Step Solution:

Step 1: Calculate the mass of pure $CaCO_3$ in the sample.
$\Rightarrow \frac{80 \times 400}{100}=320 \mathrm{~g} \text { of } \mathrm{CaCO}_3$

Step 2: Write the balanced chemical equation for the decomposition of $CaCO_3$ .
$\mathrm{CaCO}_3 \xrightarrow{\Delta} \mathrm{CaO}+\mathrm{CO}_2$

Step 3: Use stoichiometry to find the mass of CaO produced.
$100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ contains $=56 \mathrm{~g}$ of $\mathrm{CaO}$
$\begin{aligned} & \ 320 \text { of } \mathrm{CaCO}_3 \rightarrow \frac{56 \times 320}{100}=179.2 \mathrm{~g} \text { of } \mathrm{CaO}\end{aligned}$

Why other options are wrong:
- Option B: Incorrect calculation. The decimal point is misplaced.
- Option C: Incorrect calculation. The value is too large.
- Option D: Incorrect calculation. The value is too small.

Key Concepts:
- Stoichiometry
- Percentage purity
- Decomposition reactions

Memory Aids:
- Remember to account for the percentage purity when performing stoichiometric calculations.

Quick Tips:
- Always start with a balanced chemical equation.

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Application +4 | -1

When

{HNO}_3

is converted into

NH_3,

the equivalent weight of

{HNO}_3

will be:

\left({M}=\right.

molecular weight of

\left.{HNO}_3\right)

View answer and explanation

Correct option: D

{M} / 8

Option D (M/8) is correct because the equivalent weight is calculated by dividing the molecular weight by the change in oxidation number, which is 8 in this conversion.

Solution approach

Correct Answer + Core Reason: M/8. The equivalent weight is calculated by dividing the molecular weight by the change in oxidation number.

Step-by-Step Solution:

Step 1: Determine the oxidation state of nitrogen in $HNO_3$ :
$HNO_3$ : H(+1) + N(x) + 3O(-2) = 0
1 + x - 6 = 0\nx = +5

Step 2: Determine the oxidation state of nitrogen in $NH_3$ :
$NH_3$ : N(y) + 3H(+1) = 0\ny + 3 = 0\ny = -3

Step 3: Calculate the change in oxidation number:\nChange in oxidation number = |(+5) - (-3)| = 8

Step 4: Calculate the equivalent weight:\nEquivalent weight = $\frac{\text{Molecular weight}}{\text{Change in oxidation number}} = \frac{M}{8}$

Step 5: Final Answer: M/8

Why other options are wrong:
- Option A (M/2): This is incorrect because it likely involves an incorrect calculation of the change in oxidation number.
- Option B (M/1): This is incorrect because it likely involves an incorrect calculation of the change in oxidation number.
- Option C (M/6): This is incorrect because it likely involves an incorrect calculation of the change in oxidation number.

Key Concepts:
- Equivalent weight
- Oxidation states

Memory Aids:
- Remember the formula: Equivalent weight = $\frac{\text{Molecular weight}}{\text{Change in oxidation number}}$

Quick Tips:
- Always determine the oxidation states correctly.
- Remember to calculate the change in oxidation number accurately.

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Application +4 | -1

If

{N_A}

is Avogadro’s number, then number of valence electrons in 4.2 g of nitride ions (

{N^{3 - }}

) is

View answer and explanation

Correct option: A

2.4\,{N_A}\,

Option A (

2.4\,{N_A}\,

) is correct because the number of valence electrons in 4.2 g of nitride ions is calculated based on Avogadro's number and the number of valence electrons per ion.

Solution approach

Correct Answer + Core Reason: $2.4\,{N_A}\,$ . The number of valence electrons is calculated based on the mass of nitride ions and Avogadro's number.

Step-by-Step Solution:

Step 1: Determine the number of moles of nitride ions ( ${N^{3 - }}$ ) in 4.2 g.

Step 2: Calculate the molar mass of ${N^{3 - }}$ which is approximately 14 g/mol.

Step 3: Calculate the number of moles: $moles = \frac{{mass}}{{molar\,mass}} = \frac{{4.2}}{{14}} = 0.3\,mol$

Step 4: Each nitride ion ( ${N^{3 - }}$ ) has 8 valence electrons (5 from nitrogen + 3 from the 3- charge).

Step 5: Calculate the total number of valence electrons: $0.3\,mol \times 8 \times {N_A} = 2.4\,{N_A}$

Why other options are wrong:
- **Option B ( $4.2\,{N_A}$ ): This is incorrect. It likely results from not considering the number of valence electrons per nitride ion or incorrect calculation of moles.
- Option C ( $1.6\,{N_A}$ ): This is incorrect. It might be obtained by using a wrong number of valence electrons per ion or an incorrect mole calculation.
- Option D ( $3.2\,{N_A}$ ): This is incorrect. It likely involves a mistake in calculating the number of moles or the number of valence electrons per ion.

Key Concepts:
- Moles calculation
- Avogadro's number
- Valence electrons

Memory Aids:
- Remember the formula: moles = mass / molar mass
- Nitrogen has 5 valence electrons, and the 3- charge adds 3 more.

Quick Tips:**
- Always calculate the number of moles first.
- Pay attention to the charge of the ion and its effect on valence electrons.

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Analyzing +4 | -1

In the reaction,

2 \mathrm{Al(s})+6 \mathrm{HCl(a q}) \rightarrow 2 \mathrm{Al^{3+}(a q})+6 \mathrm{Cl^{-}(a q})+3 \mathrm{H_2(g})

View answer and explanation

Correct option: D

11.2 \mathrm{~L} \mathrm{~H}_2(\mathrm{g})

at STP is produced for every mole

\mathrm{HCl(a q})

consumed

Option D (

11.2 \mathrm{~L} \mathrm{~H}_2(\mathrm{g})

at STP is produced for every mole

\mathrm{HCl(a q})

consumed) is correct because the stoichiometry of the reaction dictates this relationship.

Solution approach

Correct Answer + Core Reason: 11.2 L H₂(g) at STP is produced for every mole HCl(aq) consumed because of the stoichiometry of the balanced equation.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2 \mathrm{Al(s)} + 6 \mathrm{HCl(aq)} \rightarrow 2 \mathrm{Al^{3+}(aq)} + 6 \mathrm{Cl^{-}(aq)} + 3 \mathrm{H_2(g)}$

Step 2: Determine the mole ratio between HCl and H₂. From the balanced equation, 6 moles of HCl produce 3 moles of H₂.

Step 3: Calculate the volume of H₂ produced by 6 moles of HCl at STP. 3 moles of H₂ at STP occupy a volume of $3 \times 22.4 \mathrm{~L} = 67.2 \mathrm{~L}$ .

Step 4: Calculate the volume of H₂ produced by 1 mole of HCl at STP. Since 6 moles of HCl produce 67.2 L of H₂, 1 mole of HCl produces $\frac{{67.2 \mathrm{~L}}}{{6}} = 11.2 \mathrm{~L}$ .

Step 5: Final Answer: 11.2 L of H₂ is produced at STP for every mole of HCl consumed.

Why other options are wrong:
- Option A: This is incorrect because the mole ratio between HCl and H₂ is 6:3, which simplifies to 2:1. Therefore, 6 L of HCl consumed would produce 3 L of H₂ only if the molar volume was 1 L/mol, which is not the case at STP.
- Option B: This is incorrect because 2 moles of Al produce 3 moles of H₂. Therefore, 1 mole of Al produces 1.5 moles of H₂. At STP, 1.5 moles of H₂ occupies a volume of $1.5 \times 22.4 \mathrm{~L} = 33.6 \mathrm{~L}$ . However, this is only true at STP; the volume depends on temperature and pressure.
- Option C: This is incorrect because 2 moles of Al produce 3 moles of H₂. Therefore, 1 mole of Al produces 1.5 moles of H₂. At STP, 1.5 moles of H₂ occupies a volume of $1.5 \times 22.4 \mathrm{~L} = 33.6 \mathrm{~L}$ , not 67.2 L.
- Option D: This is correct as shown in the step-by-step solution.

Key Concepts:
- Stoichiometry
- Mole concept
- Standard Temperature and Pressure (STP)
- Molar volume of a gas at STP (22.4 L/mol)

Memory Aids:
- Remember the molar volume of a gas at STP: 22.4 L/mol
- Use the balanced equation to determine mole ratios.

Quick Tips:
- Always start with a balanced chemical equation.
- Pay attention to the units and make sure they are consistent.
- Remember that the volume of a gas depends on temperature and pressure, unless specified at STP.

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Application +4 | -1

When 3.0 litre solution of normality

\mathrm{N}

is mixed with 5.0 litre of

4 \mathrm{M} \mathrm{HCl}

, then the resultant solution has the normality 10 . Find the value of

\mathrm{N}

.

View answer and explanation

Correct option: B

20

Option B (20) is correct because it accurately calculates the normality N using the mixing equation N1V1 + N2V2 = NV.

Solution approach

Correct Answer + Core Reason: 20. The value of N is calculated using the normality equation for mixing solutions.

Step-by-Step Solution:

Step 1: State the given information:\nV1 = 3.0 L, N1 = N\nV2 = 5.0 L, N2 = 4 M (which is equivalent to 4 N for HCl)\nFinal volume, V = V1 + V2 = 3.0 L + 5.0 L = 8.0 L\nFinal normality, N = 10

Step 2: Apply the mixing equation: $N_1V_1 + N_2V_2 = NV$

Step 3: Substitute the given values: $N \times 3.0 + 4 \times 5.0 = 10 \times 8$

Step 4: Simplify the equation: $3N + 20 = 80$

Step 5: Solve for N: $3N = 60$
$N = \frac{{60}}{{3}} = 20$

Why other options are wrong:
- Option A (10): This is incorrect because it doesn't correctly apply the mixing equation. It might be a result of incorrectly subtracting or dividing the values.
- Option C (15): This is incorrect due to an error in the algebraic manipulation of the mixing equation. It's likely a mistake in isolating N.
- Option D (25): This is incorrect because it involves a calculation error when solving for N in the mixing equation.

Key Concepts:
- Normality
- Mixing of solutions
- Normality equation: $N_1V_1 + N_2V_2 = NV$

Memory Aids:
- Remember that Molarity of HCl is equal to its Normality.

Quick Tips:
- Always ensure the units are consistent before applying the mixing equation.
- Remember to calculate the final volume correctly when mixing solutions.

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Application +4 | -1

A sample of clay contains

50 \%

silica and

10 \%

water. The sample is partially dried by which it loses

8 \mathrm{~g}

water. If the percentage of silica in the partially dried clay is 52, then what is the percentage of water in the partially dried clay?

View answer and explanation

Correct option: B

6.4 \%

Option B (6.4%) is correct because the percentage of water is calculated based on the conservation of silica mass before and after drying.

Solution approach

Correct Answer + Core Reason: 6.4%. The percentage of water is determined by the constant ratio of silica to other components.

Step-by-Step Solution:

Step 1: Define the initial mass of clay as 'm'. Initial mass of silica = 0.5m, Initial mass of water = 0.1m.

Step 2: After drying, the mass of water lost is 8g. Let the final mass of water be x.

Step 3: The mass of silica remains constant during drying. The mass of components other than water initially is 0.9m.

Step 4: The mass of the partially dried clay is m - 8.

Step 5: The mass of silica in the partially dried clay is 0.5m = 0.52(m-8).

Step 6: Solve for m: 0.5m = 0.52m - 4.16 => 0.02m = 4.16 => m = 208g.

Step 7: Initial mass of water = 0.1m = 0.1 * 208 = 20.8g. Final mass of water = 20.8 - 8 = 12.8g.

Step 8: Mass of partially dried clay = 208 - 8 = 200g.

Step 9: Percentage of water in the partially dried clay = (12.8/200) * 100 = 6.4%.

Why other options are wrong:
- Option A (2.0%): This is incorrect as it likely involves a miscalculation of the mass balance after drying.
- Option C (10.4%): This is incorrect as it likely involves a miscalculation of the mass balance after drying.
- Option D (2.4%): This is incorrect as it likely involves a miscalculation of the mass balance after drying.

Key Concepts:
- Mass percentage
- Mass balance

Memory Aids:
- Mass of solute remains constant during dilution/drying.

Quick Tips:
- Set up a mass balance equation to track the changes in mass.
- Ensure the percentages add up to 100%.

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Understanding +4 | -1

A mixture contains two or more substances in ________ which are called its ________.

View answer and explanation

Correct option: C

any ratio, components

Option C (any ratio, components) is correct because a mixture is defined as a combination of two or more substances (components) in any proportion, where each substance retains its individual properties.

Solution approach

Correct Answer + Core Reason: any ratio, components. Mixtures are composed of substances in any ratio, called components.

Step-by-Step Solution:

Step 1: Define a mixture.

Step 2: A mixture is a substance comprising two or more components that are not chemically combined.

Step 3: The components of a mixture can be present in any ratio.

Step 4: The components of a mixture retain their individual properties.

Step 5: Therefore, a mixture contains two or more substances in any ratio, which are called its components.

Why other options are wrong:
- Option A: Compounds have a fixed ratio, not mixtures.
- Option B: Elements are pure substances, and mixtures can contain compounds as well as elements. Also, mixtures don't have fixed ratios.
- Option D: Elements are pure substances, and mixtures can contain compounds as well as elements.

Key Concepts:
- Mixture
- Component
- Compound
- Element

Memory Aids:
- Remember that mixtures can contain any ratio of components, while compounds have a fixed ratio.

Quick Tips:
- Understand the difference between mixtures and compounds.

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Application +4 | -1

34 g of hydrogen peroxide is present in 1120 mL of solution. This solution is called:

View answer and explanation

Correct option: A

10 vol solution

Option A (10 vol solution) is correct because 34 g of H2O2 in 1120 mL of solution produces 10 mL of O2 per mL of solution.

Solution approach

Correct Answer + Core Reason: 10 vol solution. The volume strength of H2O2 is related to the volume of oxygen released.

Step-by-Step Solution:

Step 1: Determine the amount of ${\mathrm{H}_2}{\mathrm{O}_2}$ in 1 mL of solution: $= \frac{{34}}{{1120}}\,\mathrm{g}$

Step 2: Recall that 34 g of ${\mathrm{H}_2}{\mathrm{O}_2}$ gives 11.2 litre ${\mathrm{O}_2}$ at STP.

Step 3: Calculate the volume of ${\mathrm{O}_2}$ produced by $\frac{{34}}{{1120}}\,\mathrm{g}\,{\mathrm{H}_2}{\mathrm{O}_2}$ : $\frac{{34}}{{1120}}\,\mathrm{g}\,{\mathrm{H}_2}{\mathrm{O}_2} = \frac{{11.2 \times 34}}{{1120 \times 34}}\,\mathrm{litre}\,{\mathrm{O}_2}$

Step 4: Simplify the expression: $= \frac{1}{{100}}\,\mathrm{litre}\,{\mathrm{O}_2} = 10\,\mathrm{mL}\,{\mathrm{O}_2}$

Step 5: Therefore, 1 mL of the solution produces 10 mL of ${\mathrm{O}_2}$ , which means it is a 10 volume solution.

Why other options are wrong:
- Option B: A 20 vol solution would mean 1 mL of solution produces 20 mL of oxygen, which is not the case here.
- Option C: A 30 vol solution would mean 1 mL of solution produces 30 mL of oxygen, which is not the case here.
- Option D: A 32 vol solution would mean 1 mL of solution produces 32 mL of oxygen, which is not the case here.

Key Concepts:
- Volume strength of hydrogen peroxide
- Decomposition of hydrogen peroxide
- Stoichiometry

Memory Aids:
- Volume strength is the volume of oxygen released by 1 volume of H2O2 solution.

Quick Tips:
- Remember the relationship between mass of H2O2 and volume of oxygen released at STP.

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Understanding +4 | -1

Assertion:

\mathrm{NaNO}_3

has no definite molecule.
Reason: Its formula mass is 85.

View answer and explanation

Correct option: D

Both Assertion and Reason are correct but Reason is not correct explanation of Assertion

Option D is correct because while both the assertion and reason are true, the reason (formula mass) does not explain why NaNO3 lacks a definite molecule (lattice structure).

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct but Reason is not correct explanation of Assertion. $\mathrm{NaNO}_3$ forms a crystal lattice, and its formula mass doesn't explain this.

Step-by-Step Solution:

Step 1: Analyze the assertion: $\mathrm{NaNO}_3$ has no definite molecule. This is true because it exists as a crystal lattice.

Step 2: Analyze the reason: Its formula mass is 85. This is also true; the molar mass of $\mathrm{NaNO}_3$ is approximately 85 g/mol.

Step 3: Determine if the reason explains the assertion. The formula mass does not explain why $\mathrm{NaNO}_3$ exists as a lattice.

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

Key Concepts:
- Crystal lattice
- Formula mass
- Ionic compounds

Memory Aids:
- Ionic compounds form lattices, not discrete molecules.

Quick Tips:
- Consider the structure of the compound when analyzing assertions about its properties.

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Application +4 | -1

10 g of metal oxide contains 4 g of oxygen. The equivalent weight of the metal carbonate is

View answer and explanation

Correct option: A

42

Option A (42) is correct because the equivalent weight of the metal carbonate is calculated by first finding the equivalent weight of the metal and then adding the equivalent weight of the carbonate ion.

Solution approach

Correct Answer + Core Reason: 42. The equivalent weight of the metal carbonate is calculated by first finding the equivalent weight of the metal and then adding the equivalent weight of the carbonate ion.

Step-by-Step Solution:

Step 1: Given: 10 g of metal oxide contains 4 g of oxygen. Therefore, the mass of the metal is 10 - 4 = 6 g.

Step 2: Calculate the equivalent weight of the metal: $\frac{\text{Equivalent weight of metal}}{\text{Equivalent weight of oxygen}} = \frac{\text{Mass of metal}}{\text{Mass of oxygen}}$
⇒ $\frac{E}{8} = \frac{6}{4}$ ⇒ $E = \frac{6 \times 8}{4} = 12$

Step 3: Calculate the equivalent weight of the carbonate ion ( $CO_3^{2-}$ ): Equivalent weight = $\frac{\text{Molar mass}}{\text{Valency}} = \frac{60}{2} = 30$

Step 4: Calculate the equivalent weight of the metal carbonate: Equivalent weight = Equivalent weight of metal + Equivalent weight of carbonate = 12 + 30 = 42.

Why other options are wrong:
- Option B (24): This is incorrect due to a calculation error. It might be obtained by incorrectly adding the equivalent weights or by not considering the carbonate ion.
- Option C (12): This is incorrect because it only calculates the equivalent weight of the metal and doesn't add the equivalent weight of the carbonate ion.
- Option D (50): This is incorrect due to a calculation error. There is no clear logical error that would lead to this value.

Key Concepts:
- Equivalent weight
- Equivalent weight of metal oxide
- Equivalent weight of metal carbonate

Memory Aids:
- Equivalent weight = $\frac{\text{Molar mass}}{\text{Valency}}$

Quick Tips:
- Remember to consider the valency when calculating equivalent weights.
- Ensure you add the equivalent weights of all components in the compound.

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Understanding +4 | -1

The molar solution of

H_2SO_4

is equal to :

View answer and explanation

Correct option: A

N/2 solution

Option A (N/2 solution) is correct because the molarity of H2SO4 is half its normality since it is a dibasic acid with two replaceable hydrogen atoms.

Solution approach

Correct Answer + Core Reason: N/2 solution. Molarity is half the normality for dibasic acids like H2SO4.

Step-by-Step Solution:

Step 1: Define molarity (M) and normality (N):\nMolarity = moles of solute / liters of solution\nNormality = gram equivalent weight of solute / liters of solution

Step 2: Relate molarity and normality for acids: Normality = Molarity x Basicity

Step 3: Determine the basicity of H2SO4: H2SO4 is a dibasic acid, so its basicity is 2.

Step 4: Calculate the molarity of a 1 N H2SO4 solution: Molarity = Normality / Basicity = 1 N / 2 = 0.5 M = N/2

Why other options are wrong:
- Option B (N solution): This would be the normality, not the molarity.
- Option C (2N solution): This is twice the normality, not the molarity.
- Option D (3N solution): This is incorrect; the relationship is N/2.

Key Concepts:
- Molarity
- Normality
- Basicity of acids

Memory Aids:
- Normality = Molarity x n-factor (acidity/basicity)

Quick Tips:
- Remember to consider the basicity or acidity of the compound when converting between molarity and normality.

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Analyzing +4 | -1

Assertion (A) :

20 \mathrm{g}

of

50 \%

pure

\mathrm{CaCO}_3

on heating gave 2.24 L

\mathrm{CO}_2

at STP and

15.6 \mathrm{~g}

non-volatile residue.

\mathrm{CaCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_2

Reason (R): Pure

\mathrm{CaCO}_3

is

10 \mathrm{~g}

which gave

4.4 \mathrm{~g}

\mathrm{CO}_2(=0.1 \mathrm{~mol}=2.24 \mathrm{~L}

at STP

)

and

15.6 \mathrm{~g}

residue is left.

View answer and explanation

Correct option: A

Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

.

The correct answer is A because both the assertion and the reason are true, and the reason correctly explains why the given amount of impure CaCO3 produces 2.24 L of CO2 at STP.

Solution approach

Correct Answer + Core Reason: Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$ . The reason correctly explains the assertion.

Step-by-Step Solution:

Step 1: Calculate the mass of pure $\mathrm{CaCO}_3$ in the sample.
Mass of pure $\mathrm{CaCO}_3 = 20 \text{{ g}} \times 50\% = 10 \text{{ g}}$

Step 2: Calculate the moles of pure $\mathrm{CaCO}_3$ .
Moles of $\mathrm{CaCO}_3 = \frac{{10 \text{{ g}}}}{{100 \text{{ g/mol}}}} = 0.1 \text{{ mol}}$

Step 3: Write the balanced chemical equation for the decomposition of $\mathrm{CaCO}_3$ .
$\mathrm{CaCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{CaO} + \mathrm{CO}_2$

Step 4: Calculate the moles of $\mathrm{CO}_2$ produced.
From the balanced equation, 1 mole of $\mathrm{CaCO}_3$ produces 1 mole of $\mathrm{CO}_2$ . Therefore, 0.1 mol of $\mathrm{CaCO}_3$ produces 0.1 mol of $\mathrm{CO}_2$ .

Step 5: Calculate the volume of $\mathrm{CO}_2$ produced at STP.
Volume of $\mathrm{CO}_2 = 0.1 \text{{ mol}} \times 22.4 \text{{ L/mol}} = 2.24 \text{{ L}}$

Step 6: Calculate the mass of $\mathrm{CaO}$ produced.
Moles of $\mathrm{CaO} = 0.1 \text{{ mol}}$
Mass of $\mathrm{CaO} = 0.1 \text{{ mol}} \times 56 \text{{ g/mol}} = 5.6 \text{{ g}}$

Step 7: Calculate the mass of non-volatile residue.
Mass of non-volatile residue = Mass of $\mathrm{CaO}$ + Mass of impurities = 5.6 g + (20 g - 10 g) = 5.6 g + 10 g = 15.6 g

Step 8: Verify the assertion and reason.
- Assertion: 20 g of 50% pure $\mathrm{CaCO}_3$ on heating gave 2.24 L $\mathrm{CO}_2$ at STP and 15.6 g non-volatile residue. (TRUE)
- Reason: Pure $\mathrm{CaCO}_3$ is 10 g which gave 4.4 g $\mathrm{CO}_2$ (=0.1 mol = 2.24 L at STP) and 15.6 g residue is left. (TRUE and explains the assertion)

Why other options are wrong:
- Option B: The reason is the correct explanation of the assertion.
- Option C: The reason is true.
- Option D: Both assertion and reason are true.

Key Concepts:
- Stoichiometry
- Molar mass
- STP conditions
- Decomposition reactions

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

Quick Tips:
- Always start by calculating the amount of pure substance in the sample.
- Use the balanced chemical equation to determine the stoichiometric relationships.

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Application +4 | -1

0.2

g of a sample of

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

required

10

mL of

N\mathrm{KMn}{\mathrm{O}_4}

in a titration in the presence of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}\mathrm{.}

. Purity of

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

is :

View answer and explanation

Correct option: B

85\%

Option B (

85\%

) is correct because the purity is calculated by determining the mass of H2O2 that reacted with KMnO4 and then finding the percentage of that mass in the original sample.

Solution approach

Correct Answer + Core Reason: $85\%$ . The purity is calculated based on the stoichiometry of the reaction.

Step-by-Step Solution:

Step 1: Equate the milliequivalents of $H_2O_2$ and $KMnO_4$ : $Meq. of H_2O_2 = Meq. of KMnO_4$ .

Step 2: Calculate the equivalent weight of $H_2O_2$ : Equivalent weight = $\frac{Molecular weight}{n-factor} = \frac{34}{2} = 17$ .

Step 3: Use the formula: $\frac{w}{Equivalent weight} \times 1000 = Volume \times Normality$ .

Step 4: Substitute the given values: $\frac{w}{17} \times 1000 = 10 \times 1$ .

Step 5: Solve for w: $w = \frac{10 \times 17}{1000} = 0.17 \text{{ g}}$ .

Step 6: Calculate the percentage purity: $Purity = \frac{Mass of H_2O_2}{Mass of sample} \times 100 = \frac{0.17}{0.2} \times 100 = 85\%$ .

Why other options are wrong:
- **Option A ( $25\%$ ): This is incorrect. It might be obtained by using the wrong n-factor or incorrect calculation.
- Option C ( $65\%$ ): This is incorrect and doesn't relate to the correct calculation.
- Option D ( $95\%$ ): This is incorrect and doesn't relate to the correct calculation.

Key Concepts:
- Redox titrations
- Equivalent weight
- Percentage purity

Memory Aids:
- Remember to balance the redox reaction to find the correct n-factor.

Quick Tips:**
- Always equate the milliequivalents of the reactants in a titration.

225 Medium

225 of 272

Application +4 | -1

The moles of

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

are required to react completely with

2.4 \mathrm{~mol}

of

\mathrm{CuS}

in hot acidic medium

View answer and explanation

Correct option: B

3.2

Option B (3.2) is correct because the number of moles of K2Cr2O7 required is calculated using the n-factor method, equating the equivalents of K2Cr2O7 and CuS.

Solution approach

Correct Answer + Core Reason: 3.2. The moles of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ are calculated by equating the equivalents of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ and $\mathrm{CuS}$ .

Step-by-Step Solution:

Step 1: Identify the oxidation states of Cr in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ and Cu and S in CuS.

Step 2: Determine the n-factor for each compound.\nFor $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ , Cr goes from +6 to +3, so the n-factor is 2 * 3 = 6.\nFor CuS, Cu goes from +2 to +0 and S goes from -2 to +6, so the n-factor is 2 + 8 = 8.

Step 3: Use the equation $(\mathrm{n} \times \mathrm{nf})_{\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}=(\mathrm{n} \times \mathrm{nf})_{\mathrm{CuS}}$

Step 4: Substitute the given values: $\mathrm{n} \times 6 = 2.4 \times 8$

Step 5: Solve for n: $\mathrm{n} = \frac{{2.4 \times 8}}{{6}} = 3.2$

Step 6: Final Answer: 3.2

Why other options are wrong:
- Option A (1.8): This is incorrect because it uses a wrong n-factor or an incorrect setup of the equation.
- Option C (1.6): This is incorrect because it uses a wrong n-factor or an incorrect setup of the equation.
- Option D (6.4): This is incorrect because it uses a wrong n-factor or an incorrect setup of the equation.

Key Concepts:
- Redox reactions
- n-factor calculation
- Stoichiometry

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

Quick Tips:
- Always balance the redox reaction to determine the n-factors correctly.
- Remember to consider the number of atoms when calculating the n-factor.

226 Medium

226 of 272

Analyzing +4 | -1

Assertion: Components of a homogeneous mixture cannot be separated by using physical methods.

Reason: Composition of homogeneous mixture is uniform throughout as the components react to form a single compound.

View answer and explanation

Correct option: D

If both assertion and reason is false

Option D (If both assertion and reason is false) is correct because components of a homogeneous mixture can be separated by physical methods, and the components do not react to form a single compound.

Solution approach

Correct Answer + Core Reason: If both assertion and reason is false. Homogeneous mixtures can be separated by physical methods, and their components do not react to form a single compound.

Step-by-Step Solution:

Step 1: Define a homogeneous mixture: A mixture where the components are uniformly distributed throughout.

Step 2: Consider separation techniques: Physical methods like distillation, evaporation, and chromatography can separate homogeneous mixtures.

Step 3: Analyze the reason: Components of a homogeneous mixture do not chemically react to form a single compound; they remain as individual substances.

Step 4: Conclude that both the assertion and the reason are false.

Why other options are wrong:
- Option A: Both assertion and reason are false, so this option is wrong.
- Option B: Both assertion and reason are false, so this option is wrong.
- Option C: The assertion is false, so this option is wrong.

Key Concepts:
- Homogeneous mixtures
- Physical separation methods (distillation, evaporation, chromatography)
- Chemical reactions vs. physical mixing

Memory Aids:
- Homogeneous = Uniform = Separable by physical means

Quick Tips:
- Remember that mixing is a physical process unless a chemical reaction occurs.

227 Medium

227 of 272

Application +4 | -1

What volume of 0.1

M\mathrm{KMn}{\mathrm{O}_4}

is needed to oxidise 100 mg of

\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4}

in acidic solution?

View answer and explanation

Correct option: A

4.1 mL

Option A (4.1 mL) is correct because the volume of KMnO4 needed is calculated by equating the milliequivalents of KMnO4 and FeC2O4.

Solution approach

Correct Answer + Core Reason: 4.1 mL. The volume of KMnO4 is calculated by equating milliequivalents.

Step-by-Step Solution:

Step 1: Write the balanced equation for the oxidation of FeC2O4 by KMnO4 in acidic solution (this step is not strictly necessary for solving using equivalents, but helpful for understanding the stoichiometry):
$5\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4} + 3\mathrm{KMn}{\mathrm{O}_4} + 24{\mathrm{H}^ + } \to 5\mathrm{F}{\mathrm{e}^{3 + }} + 3\mathrm{M}{\mathrm{n}^{2 + }} + 10\mathrm{C}{\mathrm{O}_2} + 3{\mathrm{K}^ + } + 12{\mathrm{H}_2}\mathrm{O}$

Step 2: Determine the n-factor for FeC2O4: $\mathrm{F}{\mathrm{e}^{2 + }}\mathrm{C}_2^{2 + }{\mathrm{O}_4} \to \mathrm{F}{\mathrm{e}^{3 + }} + 2{\mathrm{C}^{4 + }}{\mathrm{O}_2} + 3e^-$ (Fe goes from +2 to +3, and each C goes from +3 to +4, so total 3 electrons are released). Therefore, n-factor = 3.

Step 3: Determine the n-factor for KMnO4: Mn goes from +7 in KMnO4 to +2 in Mn²⁺, so n-factor = 5.

Step 4: Equate the milliequivalents of KMnO4 and FeC2O4: $\mathrm{Meq}\mathrm{. of KMn}{\mathrm{O}_4} = \mathrm{Meq}\mathrm{. of Fe}{\mathrm{C}_2}{\mathrm{O}_4}$
$0.1 \times 5 \times V = \frac{{100 \times 10^{-3}}}{{144/3}} \times 1000$ , where V is the volume of KMnO4 in mL and 144 is the molar mass of FeC2O4.

Step 5: Solve for V: $0.5V = \frac{{100}}{{144}} \times 3 = \frac{{300}}{{144}}$
$V = \frac{{300}}{{144 \times 0.5}} = \frac{{300}}{{72}} = 4.166 \approx 4.1 \text{{ mL}}$

Why other options are wrong:
- Option B (8.2 mL): Incorrect calculation of n-factor or misapplication of the milliequivalent concept.
- Option C (10.2 mL): Incorrect calculation of n-factor or misapplication of the milliequivalent concept.
- Option D (4.6 mL): Incorrect calculation of n-factor or misapplication of the milliequivalent concept.

Key Concepts:
- Redox titrations
- Milliequivalents
- n-factor (number of electrons transferred)

Memory Aids:
- Normality = Molarity × n-factor
- Meq = Normality × Volume (in mL)

Quick Tips:
- Always determine the n-factors correctly for both the oxidizing and reducing agents.
- Remember to convert mass to milliequivalents using the equivalent weight.

228 Medium

228 of 272

Analyzing +4 | -1

0.1

mole

_3 \mathrm{PO}_x

is completely neutralised by

5.6 \mathrm{~g} \mathrm{~KOH}

then the true statement is:

View answer and explanation

Correct option: C

x=2

and given acid does not form acid salt

The correct answer is C (

x=2

and given acid does not form acid salt) because the neutralization reaction requires 0.1 mole of KOH to neutralize 0.1 mole of H3POx, indicating a monobasic acid.

Solution approach

Correct Answer + Core Reason: $x=2$ and given acid does not form acid salt. This is because 0.1 mole of the acid requires 0.1 mole of KOH for complete neutralization, indicating a monobasic acid, H3PO2.

Step-by-Step Solution:

Step 1: Determine the molar mass of KOH.\nMolar mass of KOH = 39 (K) + 16 (O) + 1 (H) = 56 g/mol

Step 2: Calculate the number of moles of KOH used.\nMoles of KOH = mass / molar mass = 5.6 g / 56 g/mol = 0.1 mol

Step 3: Write the general neutralization reaction for H3POx with KOH.\nH3POx + n KOH -> KnH3-nPOx + n H2O, where n is the basicity of the acid.

Step 4: Determine the basicity (n) of the acid.\nSince 0.1 mol of H3POx is neutralized by 0.1 mol of KOH, the basicity (n) is 1.

Step 5: Determine the value of x.\nIf n = 1, the acid is H3PO2. Therefore, x = 2.

Step 6: Determine if the acid forms acid salts.\nSince H3PO2 is a monobasic acid, it does not form acid salts.

Why other options are wrong:
- Option A: If x=3, the acid is H3PO3, which is dibasic and would require 0.2 moles of KOH for neutralization. This contradicts the given information.
- Option B: If x=4, the acid is H3PO4, which is tribasic and would require 0.3 moles of KOH for neutralization. This contradicts the given information. Also, H3PO4 has P-OH linkages, but no P-H linkages.
- Option D: This option is incorrect because options A and B are incorrect.

Key Concepts:
- Neutralization reactions
- Basicity of acids
- Molar mass calculations

Memory Aids:
- Monobasic acids have one replaceable hydrogen atom.
- Dibasic acids have two replaceable hydrogen atoms.
- Tribasic acids have three replaceable hydrogen atoms.

Quick Tips:
- Always calculate the number of moles of reactants to determine the stoichiometry of the reaction.
- Remember the formulas and structures of common acids like H3PO4, H3PO3, and H3PO2.

229 Medium

229 of 272

Application +4 | -1

The percentage by volume of

\mathrm{C}_3 \mathrm{H}_8

in a mixture of

\mathrm{C}_3 \mathrm{H}_8, \mathrm{CH}_4

and

\mathrm{CO}

is

36.5.

The volume of

\mathrm{CO}_2

produced when

100 \mathrm{~ml}

of the mixture is burnt in excess of

\mathrm{O}_2

is

View answer and explanation

Correct option: B

173 \mathrm{~ml}

Option B (

173 \mathrm{~ml}

) is correct because the total volume of CO2 produced is the sum of the CO2 from propane and methane combustion.

Solution approach

Correct Answer + Core Reason: $173 \mathrm{~ml}$ . The volume of CO2 is calculated based on the percentage of propane and methane in the mixture.

Step-by-Step Solution:

Step 1: Determine the volume of C3H8 in the mixture: Volume of C3H8 = 36.5% of 100 ml = 36.5 ml.

Step 2: Determine the volume of CH4 and CO in the mixture: Volume of CH4 and CO = 100 ml - 36.5 ml = 63.5 ml.

Step 3: Write the balanced equation for the combustion of C3H8: $\mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2} \mathrm{O}$ . From this, 1 volume of C3H8 produces 3 volumes of CO2.

Step 4: Write the balanced equation for the combustion of CH4: $\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2} \mathrm{O}$ . From this, 1 volume of CH4 produces 1 volume of CO2.

Step 5: Write the balanced equation for the combustion of CO: $\mathrm{CO} + \frac{1}{2}\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}$ . From this, 1 volume of CO produces 1 volume of CO2.

Step 6: Calculate the volume of CO2 produced from C3H8: Volume of CO2 from C3H8 = 3 * 36.5 ml = 109.5 ml.

Step 7: Calculate the volume of CO2 produced from CH4 and CO: Volume of CO2 from CH4 and CO = 63.5 ml (since 1 volume of each produces 1 volume of CO2).

Step 8: Calculate the total volume of CO2 produced: Total volume of CO2 = 109.5 ml + 63.5 ml = 173 ml.

Why other options are wrong:
- **Option A ( $153 \mathrm{~ml}$ ): This is incorrect because it might be calculated by not correctly accounting for the stoichiometry of the combustion reactions or by making errors in the volume calculations.
- Option C ( $193 \mathrm{~ml}$ ): This is incorrect because it might be calculated by incorrectly multiplying the volume of propane or by adding incorrect volumes.
- Option D ( $213 \mathrm{~ml}$ ): This is incorrect because it might be calculated by incorrectly multiplying the volume of propane and methane or by adding incorrect volumes.

Key Concepts:
- Stoichiometry of combustion reactions
- Volume relationships in gas reactions

Memory Aids:
- Remember that the coefficients in a balanced chemical equation represent the volume ratios for gases.

Quick Tips:**
- Always write balanced chemical equations to determine the correct stoichiometric ratios.

230 Medium

230 of 272

Analyzing +4 | -1

Statement I: Molecular mass of cane sugar

\left( {{C_{12}}{H_{22}}{O_{11}}} \right)

is

182 \, \text{amu}

Statement II: Mole of cane sugar contains

6.022 \times 10^{23}

molecules of cane sugar.

View answer and explanation

Correct option: D

Statement II is correct but statement I is incorrect

The correct answer is D (Statement II is correct but statement I is incorrect) because the molecular mass of cane sugar is 342 amu, and one mole contains Avogadro's number of molecules.

Solution approach

Correct Answer + Core Reason: Statement II is correct but statement I is incorrect. The molecular mass of cane sugar is 342 amu, and one mole contains $6.022 \times 10^{23}$ molecules.

Step-by-Step Solution:

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

Step 2: Analyze Statement I: Molecular mass of cane sugar is 182 amu. This is incorrect. The molecular mass is 342 amu.

Step 3: Analyze Statement II: Mole of cane sugar contains $6.022 \times 10^{23}$ molecules of cane sugar. This is correct. One mole of any substance contains Avogadro's number of molecules.

Step 4: Conclusion: Statement I is incorrect, and Statement II is correct.

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

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

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

Quick Tips:
- Know how to calculate molecular mass from the chemical formula.

231 Medium

231 of 272

Analyzing +4 | -1

Assertion:

\mathrm{H}_3 \mathrm{BO}_3

is monobasic Lewis acid but salt

\mathrm{Na}_3 \mathrm{BO}_3

exist.
Reason:

\mathrm{H}_3 \mathrm{BO}_3

reacts with

\mathrm{NaOH}

to give

\mathrm{Na}_3 \mathrm{BO}_3

.

View answer and explanation

Correct option: A

Assertion is correct but Reason is wrong

Option A (Assertion is correct but Reason is wrong) is correct because boric acid is a monobasic Lewis acid, and sodium borate exists, but boric acid doesn't react with NaOH to directly give Na3BO3.

Solution approach

Correct Answer + Core Reason: Assertion is correct but Reason is wrong. Boric acid is a monobasic Lewis acid, and Na3BO3 exists, but the reaction with NaOH doesn't directly give Na3BO3.

Step-by-Step Solution:

Step 1: Analyze the assertion: $\mathrm{H}_3 \mathrm{BO}_3$ is a monobasic Lewis acid, but the salt $\mathrm{Na}_3 \mathrm{BO}_3$ exists.

Step 2: Analyze the reason: $\mathrm{H}_3 \mathrm{BO}_3$ reacts with $\mathrm{NaOH}$ to give $\mathrm{Na}_3 \mathrm{BO}_3$ .

Step 3: Boric acid ( $\mathrm{H}_3 \mathrm{BO}_3$ ) acts as a Lewis acid by accepting a hydroxide ion from water: $\mathrm{B(OH)}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons [\mathrm{B(OH)}_4]^- + \mathrm{H}^+$ . It is monobasic because it can only accept one hydroxide ion.

Step 4: Sodium borate ( $\mathrm{Na}_3 \mathrm{BO}_3$ ) exists.

Step 5: Boric acid reacts with NaOH to form sodium tetraborate, $\mathrm{Na[B(OH)_4]$

232 Medium

232 of 272

Application +4 | -1

Total no. of electrons present in

48 \mathrm{~g} \mathrm{~Mg}^{2+}

are:

View answer and explanation

Correct option: C

20 N_A

Option C (

20 N_A

) is correct because it accurately calculates the total number of electrons in 48g of Mg2+ considering the electronic configuration of Mg2+.

Solution approach

Correct Answer + Core Reason: $20 N_A$ . The number of electrons is calculated based on the electronic configuration of Mg2+.

Step-by-Step Solution:

Step 1: Determine the number of electrons in a neutral Mg atom: Mg has 12 electrons.

Step 2: Determine the number of electrons in $Mg^{2+}$ ion: $Mg^{2+}$ has lost 2 electrons, so it has 10 electrons.

Step 3: Calculate the number of moles of $Mg^{2+}$ in 48 g: Molar mass of Mg is 24 g/mol, so $n = \frac{48 \text{ g}}{24 \text{ g/mol}} = 2 \text{ mol}$

Step 4: Calculate the total number of electrons: Total electrons = moles of $Mg^{2+}$ × electrons per $Mg^{2+}$ × Avogadro's number = $2 \text{ mol} \times 10 \frac{\text{electrons}}{\text{ion}} \times N_A = 20N_A$

Why other options are wrong:
- Option A: $24 N_A$ - This is incorrect because it uses the number of protons in Mg instead of the number of electrons in $Mg^{2+}$ .
- Option B: $2 N_A$ - This is incorrect because it doesn't account for the number of electrons in a single $Mg^{2+}$ ion.
- Option D: none of these - This is incorrect because option C is the correct answer.

Key Concepts:
- Electronic configuration
- Mole concept
- Avogadro's number

Memory Aids:
- Remember that cations have fewer electrons than the neutral atom.

Quick Tips:
- Determine the number of electrons in the ion correctly.
- Use the mole concept to calculate the total number of ions.

233 Medium

233 of 272

Application +4 | -1

10 \mathrm{~g} \mathrm{CaCO}_3

were dissolved in

250 \mathrm{~mL}

of

100 \mathrm{M}

\mathrm{HCl}

or the solution was boiled. What volume of

2 \mathrm{M} \mathrm{KOH}

would be required to equivalence point after boiling? Assume no change in volume during boiling.

View answer and explanation

Correct option: B

25 \mathrm{~mL}

Option B (25 mL) is correct because it is the volume of KOH required to neutralize the remaining HCl after the reaction with CaCO₃.

Solution approach

Correct Answer + Core Reason: $25 \mathrm{~mL}$ . The volume of KOH required is calculated based on the amount of HCl remaining after the reaction with CaCO3.

Step-by-Step Solution:

Step 1: Calculate the moles of $CaCO_3$ : Moles = $\frac{{10 \text{{ g}}}}{{100 \text{{ g/mol}}}} = 0.1 \text{{ mol}}$

Step 2: Calculate the moles of HCl initially: Moles = $0.250 \text{{ L}} \times 1 \text{{ M}} = 0.25 \text{{ mol}}$

Step 3: The reaction between $CaCO_3$ and $HCl$ is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$ . Therefore, 1 mole of $CaCO_3$ reacts with 2 moles of $HCl$ .

Step 4: Calculate the moles of HCl that reacted with $CaCO_3$ : Moles = $2 \times 0.1 \text{{ mol}} = 0.2 \text{{ mol}}$

Step 5: Calculate the moles of HCl remaining: Moles = $0.25 - 0.2 = 0.05 \text{{ mol}}$

Step 6: Calculate the volume of 2 M KOH required to neutralize the remaining HCl: $HCl + KOH \rightarrow KCl + H_2O$ . Therefore, 1 mole of HCl reacts with 1 mole of KOH.\nVolume of KOH = $\frac{{0.05 \text{{ mol}}}}{{2 \text{{ M}}}} = 0.025 \text{{ L}} = 25 \text{{ mL}}$

Why other options are wrong:
- **Option A (50 ~mL $):** This is incorrect. It is likely a calculation error or misunderstanding of the stoichiometry of the reaction.\n- **Option C (75 \mathrm{~mL}$ ): This is incorrect. It might be due to an incorrect calculation of the moles of HCl remaining.
- Option D (60 ~mL $):** This is incorrect. It is likely a calculation error in the formula.\n\n**Key Concepts:**\n- Stoichiometry\n- Molarity\n- Neutralization reaction\n\n**Memory Aids:**\n- Remember to balance the chemical equation.\n\n**Quick Tips:**\n- Ensure correct units are used.\n- Calculate the moles of reactants and products accurately.$

234 Medium

234 of 272

Application +4 | -1

The mole fraction of

\mathrm{CH}_3 \mathrm{OH}

in its

32 \% \,(\mathrm{w} / \mathrm{w})

aqueous solution is nearly

View answer and explanation

Correct option: C

0.21

Option C (0.21) is correct because the mole fraction is calculated by dividing the moles of methanol by the total moles of the solution.

Solution approach

Correct Answer + Core Reason: 0.21. The mole fraction is calculated by dividing the moles of methanol by the total moles of the solution.

Step-by-Step Solution:

Step 1: Assume 100 g of solution. This means 32 g of CH3OH and 68 g of H2O.

Step 2: Calculate moles of CH3OH: $n_{CH_3OH} = \frac{32 \text{{ g}}}{32 \text{{ g/mol}}} = 1 \text{{ mol}}$

Step 3: Calculate moles of H2O: $n_{H_2O} = \frac{68 \text{{ g}}}{18 \text{{ g/mol}}} = 3.77 \text{{ mol}}$

Step 4: Calculate mole fraction of CH3OH: $X_{CH_3OH} = \frac{n_{CH_3OH}}{n_{CH_3OH} + n_{H_2O}} = \frac{1}{1 + 3.77} = \frac{1}{4.77} = 0.21$

Step 5: Final Answer: 0.21

Why other options are wrong:
- Option A (0.56): This is incorrect because it likely involves an error in the mole calculation or an incorrect ratio.
- Option B (0.11): This is incorrect because it likely involves an error in the mole calculation or an incorrect ratio.
- Option D (0.38): This is incorrect because it likely involves an error in the mole calculation or an incorrect ratio.

Key Concepts:
- Mole fraction calculation
- Molar mass

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

Quick Tips:
- Always convert mass to moles before calculating mole fraction.
- Check that the sum of all mole fractions equals 1.

235 Medium

235 of 272

Application +4 | -1

5000 \mathrm{~mL}

of

0.6 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4

acid is mixed with

2 \mathrm{~L}

of 98% (w / V) of

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

(specific gravity 1.84). If the specific gravity of the resulting mixture is 1.4, what is the molarity of the mixture? (Given that the specific gravity of

0.6 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4

is 1.02 and also consider there is no loss of mass due to mixing.)

View answer and explanation

Correct option: B

3.6674 \mathrm{~M}

Option B (

3.6674 \mathrm{~M}

) is correct because it is calculated using the total moles of H2SO4 and the total volume of the mixture.

Solution approach

Correct Answer + Core Reason: $3.6674 \mathrm{~M}$ . The molarity of the mixture is calculated from the total moles of sulfuric acid and the final volume of the solution.

Step-by-Step Solution:

Step 1: Calculate the moles of $H_2SO_4$ in the first solution: $5000 \mathrm{mL} \times 0.6 \mathrm{M} = 3000 \mathrm{mmol} = 3 \mathrm{mol}$

Step 2: Calculate the mass of $H_2SO_4$ in the second solution: $2 \mathrm{L} \times 98\% (w/V) = 2000 \mathrm{mL} \times \frac{98 \mathrm{g}}{100 \mathrm{mL}} = 1960 \mathrm{g}$

Step 3: Calculate the moles of $H_2SO_4$ in the second solution: $\frac{1960 \mathrm{g}}{98 \mathrm{g/mol}} = 20 \mathrm{mol}$

Step 4: Calculate the total moles of $H_2SO_4$ : $3 \mathrm{mol} + 20 \mathrm{mol} = 23 \mathrm{mol}$

Step 5: Calculate the total mass of the two solutions:\nMass of first solution = Volume * Specific gravity = $5000 \mathrm{mL} * 1.02 \mathrm{g/mL} = 5100 \mathrm{g}$ \nMass of second solution = Volume * Specific gravity = $2000 \mathrm{mL} * 1.84 \mathrm{g/mL} = 3680 \mathrm{g}$ \nTotal mass = $5100 + 3680 = 8780 \mathrm{g}$

Step 6: Calculate the total volume of the mixture:\nVolume = Total mass / Specific gravity = $8780 \mathrm{g} / 1.4 \mathrm{g/mL} = 6271.43 \mathrm{mL} = 6.27143 \mathrm{L}$

Step 7: Calculate the molarity of the mixture:\nMolarity = Total moles / Total volume = $\frac{23 \mathrm{mol}}{6.27143 \mathrm{L}} = 3.6674 \mathrm{M}$

Why other options are wrong:
- Option A: This is incorrect due to a miscalculation in the number of moles or the final volume.
- Option C: This is incorrect due to a miscalculation in the number of moles or the final volume.
- Option D: This is incorrect due to a miscalculation in the number of moles or the final volume.

Key Concepts:
- Molarity
- Specific Gravity
- Solution Stoichiometry

Memory Aids:
- Remember that Molarity = moles of solute / volume of solution (in liters).

Quick Tips:
- Ensure you convert all volumes to the same units (e.g., liters).
- Pay attention to the units of specific gravity and concentration.

236 Medium

236 of 272

Analyzing +4 | -1

Statement I: When a hydrocarbon is burnt and the products of combustion are cooled to the original temperature and pressure, a contraction in volume occurs.

Statement II: The contraction in volume is solely due to the liquefaction of water vapours.

View answer and explanation

Correct option: C

If Statement I is CORRECT, but Statement II is INCORRECT

Option C is correct. Statement I is correct because combustion of hydrocarbons usually results in a contraction in volume. Statement II is incorrect because the contraction is due to the change in the number of moles of gaseous substances, not solely the liquefaction of water.

Solution approach

Correct Answer + Core Reason: If Statement I is CORRECT, but Statement II is INCORRECT. Contraction in volume is due to change in moles of gases, not just water liquefaction.

Step-by-Step Solution:

Step 1: Consider the general combustion reaction of a hydrocarbon: ${C}_x{H}_y + (x + \frac{y}{4}){O}_2 \rightarrow x {CO}_2 + \frac{y}{2} {H}_2 {O}$ .

Step 2: Calculate the change in the number of moles of gaseous substances. Initially, we have $1 + x + \frac{y}{4}$ moles of gas. After combustion, we have $x$ moles of $CO_2$ and $\frac{y}{2}$ moles of $H_2O$ (as gas initially). When cooled to original temperature and pressure, water condenses to liquid.

Step 3: The contraction in volume is given by the difference in the number of moles of gaseous reactants and products: $(1 + x + \frac{y}{4}) - x = 1 + \frac{y}{4}$ . This shows a contraction occurs.

Step 4: Statement I is correct because there is a contraction in volume.

Step 5: Statement II is incorrect because the contraction is not solely due to the liquefaction of water. It is due to the overall change in the number of moles of gaseous substances.

Why other options are wrong:
- Option A: If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I: Incorrect because Statement II is incorrect.
- Option B: If both statements are CORRECT, and Statement II is not the CORRECT explanation of Statement I: Incorrect because Statement II is incorrect.
- Option D: If Statement I is INCORRECT, but Statement II is CORRECT: Incorrect because Statement I is correct.

Key Concepts:
- Combustion reactions of hydrocarbons
- Stoichiometry of gaseous reactions
- Change in the number of moles of gases

Memory Aids:
- Remember that the change in the number of moles of gases affects the volume.

Quick Tips:
- Always balance the chemical equation before analyzing the change in volume.

237 Medium

237 of 272

Analyzing +4 | -1

Which of the following pairs contains equal number of atoms?

View answer and explanation

Correct option: A

11.2 cc (STP) of nitrogen and 0.015 g of nitric oxide

The correct answer is A because both 11.2 cc of nitrogen and 0.015 g of nitric oxide contain the same number of atoms.

Solution approach

Correct Answer + Core Reason: 11.2 cc (STP) of nitrogen and 0.015 g of nitric oxide because they contain the same number of atoms.

Step-by-Step Solution:

Step 1: Calculate the number of atoms in 11.2 cc of nitrogen (N2) at STP:
- Volume = 11.2 cc = 11.2 x 10-3 L
- Moles of N2 = (11.2 x 10-3 L) / (22.4 L/mol) = 0.5 x 10-3 mol
- Number of N2 molecules = (0.5 x 10-3 mol) x (6.023 x 1023 molecules/mol) = 3.0115 x 1020 molecules
- Number of N atoms = 2 x (3.0115 x 1020) = 6.023 x 1020 atoms

Step 2: Calculate the number of atoms in 0.015 g of nitric oxide (NO):
- Molar mass of NO = 14 + 16 = 30 g/mol
- Moles of NO = (0.015 g) / (30 g/mol) = 0.5 x 10-3 mol
- Number of NO molecules = (0.5 x 10-3 mol) x (6.023 x 1023 molecules/mol) = 3.0115 x 1020 molecules
- Number of atoms = 2 x (3.0115 x 1020) = 6.023 x 1020 atoms

Step 3: Compare the number of atoms: Both samples contain the same number of atoms (6.023 x 1020).

Why other options are wrong:
- Option B: 22.4 L of nitrous oxide (N2O) and 22.4 L of nitric oxide (NO) have different numbers of atoms because N2O has 3 atoms per molecule, while NO has 2.
- Option C: 1 millimole of HCl has 2 millimoles of atoms, and 0.5 millimole of H2S has 1.5 millimoles of atoms. They are not equal.
- Option D: 1 mole of H2O2 has 4 moles of atoms, and 1 mole of N2O4 has 6 moles of atoms. They are not equal.

Key Concepts:
- Mole Concept
- Avogadro's Number
- STP Conditions

Memory Aids:
- Remember that 1 mole of any gas occupies 22.4 L at STP.
- Avogadro's number (6.023 x 1023) relates moles to the number of particles.

Quick Tips:
- Always convert mass or volume to moles before calculating the number of atoms or molecules.

238 Medium

238 of 272

Analyzing +4 | -1

Maximum magnitude of work done is obtained during the hydrolysis of

1 \mathrm{~mol}

of

View answer and explanation

Correct option: A

\mathrm{Al}_4 \mathrm{C}_3

Option A (Al4C3) is correct because its hydrolysis produces the largest volume of gas (methane), resulting in the maximum magnitude of work done.

Solution approach

Correct Answer + Core Reason: $\mathrm{Al}_4 \mathrm{C}_3$ . Maximum volume of gas will be released during its hydrolysis.

Step-by-Step Solution:

Step 1: Write the hydrolysis reactions for each compound.
$\mathrm{Al}_4 \mathrm{C}_3 + 12 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Al(OH)}_3 + 3 \mathrm{CH}_4 (g)$
$\mathrm{CaC}_2 + 2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Ca(OH)}_2 + \mathrm{C}_2 \mathrm{H}_2 (g)$
$\mathrm{Be}_2 \mathrm{C} + 4 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{Be(OH)}_2 + \mathrm{CH}_4 (g)$
$\mathrm{Mg}_2 \mathrm{C}_3 + 4 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{Mg(OH)}_2 + \mathrm{C}_3 \mathrm{H}_4 (g)$

Step 2: Determine the moles of gas produced per mole of compound hydrolyzed.
$\mathrm{Al}_4 \mathrm{C}_3$ : 3 moles of $\mathrm{CH}_4$
$\mathrm{CaC}_2$ : 1 mole of $\mathrm{C}_2 \mathrm{H}_2$
$\mathrm{Be}_2 \mathrm{C}$ : 1 mole of $\mathrm{CH}_4$
$\mathrm{Mg}_2 \mathrm{C}_3$ : 1 mole of $\mathrm{C}_3 \mathrm{H}_4$

Step 3: Relate the moles of gas produced to the work done. The work done is proportional to the volume of gas produced. Since the pressure is constant, more moles of gas mean more work done.

Step 4: Conclusion: $\mathrm{Al}_4 \mathrm{C}_3$ produces the most moles of gas (3 moles of $\mathrm{CH}_4$ ) and thus has the maximum magnitude of work done during hydrolysis.

Why other options are wrong:
- **Option B ( $\mathrm{CaC}_2$ ): Produces only 1 mole of gas.
- Option C ( $\mathrm{Be}_2 \mathrm{C}$ ): Produces only 1 mole of gas.
- Option D ( $\mathrm{Mg}_2 \mathrm{C}_3$ ): Produces only 1 mole of gas.

Key Concepts:
- Hydrolysis Reactions
- Work Done by a Gas

Memory Aids:
- More gas produced = more work done.

Quick Tips:**
- Balance the hydrolysis reactions to determine the moles of gas produced.

239 Medium

239 of 272

Application +4 | -1

A compound contains 90 % C and 10 % H. The empirical formula of the compound is

View answer and explanation

Correct option: C

{C_3}{H_4}

Option C (

{C_3}{H_4}

) is correct because it represents the simplest whole-number ratio of carbon to hydrogen atoms based on the given percentages.

Solution approach

Correct Answer + Core Reason: ${C_3}{H_4}$ . The empirical formula is determined by finding the simplest whole number ratio of the elements.

Step-by-Step Solution:

Step 1: Assume we have 100g of the compound. This means we have 90g of C and 10g of H.

Step 2: Convert the mass of each element to moles using their respective atomic masses.\nMoles of C = $\frac{{90 \text{{g}}}}{{12 \text{{g/mol}}}} = 7.5 \text{{ mol}}$ \nMoles of H = $\frac{{10 \text{{g}}}}{{1 \text{{g/mol}}}} = 10 \text{{ mol}}$

Step 3: Find the simplest whole number ratio by dividing each mole value by the smallest mole value (7.5).\nRatio of C = $\frac{{7.5}}{{7.5}} = 1$ \nRatio of H = $\frac{{10}}{{7.5}} = 1.33$

Step 4: Multiply the ratio by a small integer to obtain whole numbers. In this case, multiply by 3.\nC : H = 1 : 1.33 $\Rightarrow$ 3 : 3.99 $\approx$ 3 : 4

Step 5: Write the empirical formula using the whole number ratios: ${C_3}{H_4}$

Why other options are wrong:
- **Option A ( ${C_8}{H_{10}}$ ): This ratio does not correspond to the simplest whole number ratio derived from the given percentages. The C:H ratio is 8:10 or 4:5, which doesn't match the 90% C and 10% H composition.
- Option B ( ${C_{15}}{H_{30}}$ ):** This ratio simplifies to ${C}{H_2}$ , which doesn't match the calculated ratio of C:H = 3:4. It also doesn't reflect the given percentages.
- **Option D ( ${C_5}{H_{40}}$ ): This ratio doesn't match the calculated ratio of C:H = 3:4. It also doesn't reflect the given percentages. It suggests a much higher proportion of hydrogen than is present in the compound.

Key Concepts:
- Empirical Formula
- Mole Concept
- Percentage Composition

Memory Aids:
- Percent to Mass, Mass to Mole, Divide by Small, Multiply 'til Whole.

Quick Tips:**
- Always convert percentages to masses assuming a 100g sample.
- Divide by the smallest mole value to find the simplest ratio.
- If the ratio isn't a whole number, multiply by a small integer (2, 3, 4...) to get whole numbers.

240 Medium

240 of 272

Analyzing +4 | -1

Which of the following has greatest no. of atoms

View answer and explanation

Correct option: A

1g of butane

({C_4}{H_{10}})

Option A (1g of butane

({C_4}{H_{10}})

) is correct because it has the highest number of atoms due to its lower molar mass and higher atomicity compared to the other options.

Solution approach

Correct Answer + Core Reason: 1g of butane $({C_4}{H_{10}})$ . Butane has the lowest molar mass and highest atomicity among the options.

Step-by-Step Solution:

Step 1: Define the formula: Number of atoms = (mass / molar mass) * Avogadro's number * atomicity

Step 2: Calculate for each option:
A. Butane ( $C_4H_{10}$ ): Molar mass = 58 g/mol, Atomicity = 14. Number of atoms = (1/58) * $N_A$ * 14 = 0.241 $N_A$
B. Nitrogen ( $N_2$ ): Molar mass = 28 g/mol, Atomicity = 2. Number of atoms = (1/28) * $N_A$ * 2 = 0.071 $N_A$
C. Silver (Ag): Molar mass = 108 g/mol, Atomicity = 1. Number of atoms = (1/108) * $N_A$ * 1 = 0.009 $N_A$
D. Water ( $H_2O$ ): Molar mass = 18 g/mol, Atomicity = 3. Number of atoms = (1/18) * $N_A$ * 3 = 0.167 $N_A$

Step 3: Compare the number of atoms: Butane > Water > Nitrogen > Silver

Why other options are wrong:
- **Option B (1g of Nitrogen $\left( {{N_2}} \right)$ ): Nitrogen has a higher molar mass and lower atomicity than butane, resulting in fewer atoms.
- Option C (1g of Silver $\left( {Ag} \right)$ ): Silver has a significantly higher molar mass and an atomicity of 1, resulting in the fewest atoms.
- Option D (1g of Water $\left( {{H_2}O} \right)$ ): Water has a higher molar mass and lower atomicity than butane, resulting in fewer atoms.

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

Memory Aids:**
- Number of atoms = (mass / molar mass) * $N_A$ * atomicity

Quick Tips:
- The substance with the lowest molar mass and highest atomicity for a given mass will have the greatest number of atoms.

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