25 mL of 0.50 $M{{\rm{H}}_2}{{\rm{O}}_2}$ solution is added to 50 mL of 0.20 M $M{\rm{KMn}}{{\rm{O}}_4}$ in acidic solution. Which of the following statements is true?

0.005 mole of ${\rm{KMn}}{{\rm{O}}_4}$ are left

0.010 mole of oxygen is liberated

0.030 g atom of oxygen gas is evolved

0.0025 mole ${{\rm{H}}_2}{{\rm{O}}_2}$ does not react with ${\rm{KMn}}{{\rm{O}}_4}$

SOME BASIC CONCEPTS OF CHEMISTRY MCQs for JEE CHEMISTRY — Medium, Page 7

181 Medium

181 of 272

Understanding +4 | -1

In a gaseous reaction of the type

aA + bB \to cC + dD,

which is wrong?

View answer and explanation

Correct option: C

a

g of

A

combines with

b

g of

B

to give

C

and

D

Option C (

a

g of

A

combines with

b

g of

B

to give

C

and

D

) is correct because stoichiometry in chemical reactions represents mole ratios or volume ratios (for gases), but not directly mass ratios.

Solution approach

Correct Answer + Core Reason: $a$ g of $A$ combines with $b$ g of $B$ to give $C$ and $D$ . Stoichiometry represents mole or volume ratios, not mass ratios.

Step-by-Step Solution:

Step 1: Understand stoichiometry. Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction.

Step 2: Consider the given reaction. $aA + bB \to cC + dD$ .

Step 3: Relate coefficients to moles. The coefficients (a, b, c, d) represent the number of moles of each substance involved in the reaction.

Step 4: Relate coefficients to volume (for gases). For gases, the coefficients also represent the volume ratios under the same conditions of temperature and pressure.

Step 5: Recognize that mass ratios are not directly represented by the coefficients. The coefficients do not directly represent the mass ratios because the molar masses of the substances are different.

Why other options are wrong:
- Option A: $a$ litre of $A$ combines with $b$ litre of $B$ to give $C$ and $D$ is correct for gases under the same conditions.
- Option B: $a$ mole of $A$ combines with $b$ mole of $B$ to give $C$ and $D$ is correct based on stoichiometry.
- Option D: $a$ molecules of $A$ combines with $b$ molecules of $B$ to give $C$ and $D$ is correct because the mole ratio is the same as the molecule ratio.

Key Concepts:
- Stoichiometry
- Mole ratio
- Volume ratio (for gases)
- Mass ratio

Memory Aids:
- Coefficients = moles (or volume for gases)

Quick Tips:
- Remember that stoichiometry deals with mole ratios, not mass ratios directly.

182 Medium

182 of 272

Analyzing +4 | -1

Which of the following statements illustrate the law of multiple proportions?

View answer and explanation

Correct option: C

An element forms two oxides,

\mathrm{XO}

and

\mathrm{XO}_2

containing

50 \%

and

60 \%

oxygen respectively. The ratio of masses of oxygen which combines with

1 \mathrm{~g}

of element is

2: 3

Option C is correct because it directly demonstrates the law of multiple proportions by showing that the ratio of oxygen masses combining with a fixed mass of the element is a simple whole number ratio (2:3).

Solution approach

Correct Answer + Core Reason: C. An element forms two oxides, $\mathrm{XO}$ and $\mathrm{XO}_2$ containing $50 \%$ and $60 \%$ oxygen respectively. The ratio of masses of oxygen which combines with $1 \mathrm{~g}$ of element is $2: 3$ . This illustrates the law of multiple proportions.

Step-by-Step Solution:

Step 1: Understand the Law of Multiple Proportions. This law states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Step 2: Analyze the given information for option C.\nIn XO, 50g of element combines with 50g of oxygen. Therefore, 1g of element combines with 1g of oxygen.\nIn XO2, 40g of element combines with 60g of oxygen. Therefore, 1g of element combines with 60/40 = 1.5g of oxygen.

Step 3: Calculate the ratio of oxygen masses.\nThe ratio of masses of oxygen that combine with 1g of element is 1 : 1.5, which simplifies to 2:3.

Step 4: Conclusion. Since the ratio is a simple whole number ratio (2:3), it illustrates the law of multiple proportions.

Why other options are wrong:
- Option A: This option illustrates the Law of Conservation of Mass, not the Law of Multiple Proportions. The total mass of reactants equals the total mass of products.
- Option B: This option provides percentage compositions of different compounds, which is related to the Law of Definite Proportions, not the Law of Multiple Proportions.
- Option D: This option illustrates Gay-Lussac's Law of Combining Volumes, which deals with the volumes of gases in chemical reactions, not the Law of Multiple Proportions.

Key Concepts:
- Law of Multiple Proportions
- Law of Conservation of Mass
- Law of Definite Proportions
- Gay-Lussac's Law of Combining Volumes

Memory Aids:
- Multiple Proportions: Think of multiple compounds formed from the same elements, and look for simple whole number ratios of masses.

Quick Tips:
- When checking for the Law of Multiple Proportions, ensure you are comparing different compounds formed from the *same* two elements.

183 Medium

183 of 272

Analyzing +4 | -1

In redox reaction,

{H}_2 {O}_2

oxidises

{K}_4\left[{Fe}({CN})_6\right]

into

{K}^{+}, {Fe}^{+3}, {CO}_3{ }^{-2}

and

{NO}_3^-

ions in acidic medium, than how many moles of

{H}_2 {O}_2

will react with 1mole of

{K}_4\left[{Fe}({CN})_6\right]

View answer and explanation

Correct option: D

30.5 moles

Option D is correct because the n-factor of H2O2 is 2 and the n-factor of K4[Fe(CN)6] is 61. Therefore, 30.5 moles of H2O2 will react with 1 mole of K4[Fe(CN)6].

Solution approach

Correct Answer + Core Reason: 30.5 moles. The number of moles of H2O2 required is calculated based on the n-factors of the reactants in the redox reaction.

Step-by-Step Solution:

Step 1: Determine the oxidation state changes in the reaction:\nFe in K4[Fe(CN)6] changes from +2 to +3 (oxidation).\nC in [Fe(CN)6] changes from +2 to +4 in CO32- (oxidation).\nN in [Fe(CN)6] changes from -3 to +5 in NO3- (oxidation).

Step 2: Calculate the n-factor of K4[Fe(CN)6]:\nFe: +2 to +3 (change of 1, 1 atom) = 1\nC: +2 to +4 (change of 2, 6 atoms) = 12\nN: -3 to +5 (change of 8, 6 atoms) = 48\nn-factor of K4[Fe(CN)6] = 1 + 12 + 48 = 61

Step 3: Determine the n-factor of H2O2: In acidic medium, H2O2 acts as an oxidizing agent and is reduced to H2O. The oxidation state of oxygen changes from -1 to -2. Therefore, the n-factor of H2O2 is 2.

Step 4: Apply the principle of equivalence: For a redox reaction, the number of gram equivalents of the oxidizing agent is equal to the number of gram equivalents of the reducing agent.
$\text{moles of } H_2O_2 \times n_1 = \text{moles of } K_4[Fe(CN)_6] \times n_2$

Step 5: Calculate the moles of H2O2 required to react with 1 mole of K4[Fe(CN)6]:\nmoles of H2O2 × 2 = 1 × 61\nmoles of H2O2 = 61/2 = 30.5

Step 6: Conclusion: 30.5 moles of H2O2 will react with 1 mole of K4[Fe(CN)6].

Why other options are wrong:
- Option A: Incorrect because it does not account for the complete oxidation of all elements in the complex.
- Option B: Incorrect because it does not account for the complete oxidation of all elements in the complex.
- Option C: Incorrect because it does not account for the complete oxidation of all elements in the complex.

Key Concepts:
- Redox reactions
- Oxidation state
- n-factor
- Principle of equivalence

Memory Aids:
- n-factor: The number of electrons gained or lost per molecule.

Quick Tips:
- Determine the oxidation state changes for all elements involved in the redox reaction.
- Calculate the n-factor for each reactant.
- Apply the principle of equivalence to determine the stoichiometry of the reaction.

184 Medium

184 of 272

Remembering +4 | -1

Number of gas molecules present in

1 \mathrm{~ml}

of gas at

0^{\circ} \mathrm{C}

and

1 \mathrm{~atm}

is called Loschmidt number. Its value is about

View answer and explanation

Correct option: A

2.7 \times 10^{19}

Option A (

2.7 \times 10^{19}

) is correct because the Loschmidt number, representing the number of gas molecules in 1 ml at standard conditions, is approximately

2.7 \times 10^{19}

.

Solution approach

Correct Answer + Core Reason: $2.7 \times 10^{19}$ . Loschmidt number is approximately $2.7 \times 10^{19}$ .

Step-by-Step Solution:

Step 1: Recall the definition of the Loschmidt number.

Step 2: The Loschmidt number is the number of gas molecules in 1 ml of gas at $0^{\circ} \mathrm{C}$ and 1 atm.

Step 3: The value of the Loschmidt number is approximately $2.6867774 \times 10^{19} \mathrm{cm}^{-3}$ .

Step 4: This value is approximately $2.7 \times 10^{19}$ .

Why other options are wrong:
- Option A: $2.7 \times 10^{19}$ is correct because it is the approximate value of the Loschmidt number.
- Option B: $6 \times 10^{23}$ is incorrect because it is Avogadro's number, not the Loschmidt number.
- Option C: $2.7 \times 10^{22}$ is incorrect because it is not the correct value of the Loschmidt number.
- Option D: $1.3 \times 10^{28}$ is incorrect because it is not the correct value of the Loschmidt number.

Key Concepts:
- Loschmidt number
- Avogadro's number

Memory Aids:
- Remember the approximate value of the Loschmidt number as $2.7 \times 10^{19}$ .

Quick Tips:
- Distinguish between the Loschmidt number and Avogadro's number.

185 Medium

185 of 272

Application +4 | -1

100 \mathrm{~mL}

of

2.4 \mathrm{~V} {~H}_2 {O}_2

require how many grams of

{KMnO}_4

in acidic medium for complete reaction?

View answer and explanation

Correct option: D

1.35 \mathrm{~g}

Option D (

1.35 \mathrm{~g}

) is correct because it is calculated by equating the milliequivalents of H2O2 and KMnO4 in acidic medium.

Solution approach

Correct Answer + Core Reason: $1.35 \mathrm{~g}$ . The mass of $KMnO_4$ is calculated by equating the milliequivalents of $H_2O_2$ and $KMnO_4$ .

Step-by-Step Solution:

Step 1: Calculate the normality of $H_2O_2$ using the volume strength formula: Volume strength $= 5.6 \times$ Normality. Given volume strength = 2.4.

Step 2: Normality of $H_2O_2$ , $N = \frac{2.4}{5.6} = 0.4286 \approx 0.43$

Step 3: Equate the milliequivalents of $H_2O_2$ and $KMnO_4$ : meqs. of $H_2O_2$ = meqs. of $KMnO_4$

Step 4: Calculate milliequivalents of $H_2O_2$ : $N \times V = 0.43 \times 100 = 43$

Step 5: Calculate the equivalent weight of $KMnO_4$ : Equivalent weight = $\frac{\text{Molecular weight}}{\text{Number of electrons transferred}} = \frac{158}{5} = 31.6$ (since $Mn^{+7}$ changes to $Mn^{+2}$ in acidic medium, the number of electrons transferred is 5).

Step 6: Calculate the weight of $KMnO_4$ using the milliequivalent formula: meqs. of $KMnO_4 = \frac{W \times 1000}{\text{Equivalent weight}}$ , where W is the weight of $KMnO_4$ .

Step 7: $43 = \frac{W \times 1000}{31.6}$

Step 8: Solve for W: $W = \frac{43 \times 31.6}{1000} = 1.3588 \approx 1.35 \mathrm{~g}$

Why other options are wrong:
- **Option A ( $6.6 \mathrm{~g}$ ): This is incorrect due to an error in the calculation or incorrect application of the formula.
- Option B ( $0.66 \mathrm{~g}$ ): This is incorrect due to an error in the calculation or incorrect application of the formula.
- Option C ( $0.35 \mathrm{~g}$ ): This is incorrect due to an error in the calculation or incorrect application of the formula.

Key Concepts:**
- Volume strength of $H_2O_2$
- Normality
- Milliequivalents
- Equivalent weight
- Redox reactions

Memory Aids:
- Volume strength = 5.6 x Normality
- In acidic medium, $KMnO_4$ gains 5 electrons.

Quick Tips:
- Always balance the redox reaction to find the number of electrons transferred.
- Remember to convert volume to liters or use milliequivalents consistently.

186 Medium

186 of 272

Application +4 | -1

A sample of nitric acid is

69 \%

by mass and it has a concentration of 15.44 moles per litre. Its density is

View answer and explanation

Correct option: B

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

Option B (1.41 g/cc) is correct because the density is calculated using the molarity, percentage by mass, and molar mass of nitric acid.

Solution approach

Correct Answer + Core Reason: 1.41 g/cc. Density is calculated using the molarity and percentage by mass.

Step-by-Step Solution:

Step 1: Given: Nitric acid is 69% by mass, Molarity = 15.44 moles/L.

Step 2: Assume 1 L of solution. Then, moles of $HNO_3 = 15.44$ . Mass of $HNO_3 = 15.44 \times 63 = 972.72 \text{{ g}}$ (Molar mass of $HNO_3 = 63 \text{{ g/mol}}$ ).

Step 3: Since the solution is 69% by mass, we have: $\frac{{\text{{Mass of }} HNO_3}}{{\text{{Mass of solution}}}} = 0.69$

Step 4: Mass of solution = $\frac{{\text{{Mass of }} HNO_3}}{{0.69}} = \frac{{972.72}}{{0.69}} = 1409.74 \text{{ g}}$

Step 5: Volume of solution = 1 L = 1000 mL.

Step 6: Density = $\frac{{\text{{Mass}}}}{{\text{{Volume}}}} = \frac{{1409.74 \text{{ g}}}}{{1000 \text{{ mL}}}} = 1.40974 \text{{ g/mL}} \approx 1.41 \text{{ g/cc}}$

Why other options are wrong:
- Option A (1.86 g/cc): This is incorrect. It might be calculated by incorrectly using the percentage by mass or making a calculation error.
- Option C (2.60 g/cc): This is incorrect. It might be calculated by incorrectly using the percentage by mass or making a calculation error.
- Option D (1.02 g/cc): This is incorrect. It might be calculated by incorrectly using the percentage by mass or making a calculation error.

Key Concepts:
- Molarity
- Density
- Percentage by mass

Memory Aids:
- Molarity = (Mass of solute / Molar mass of solute) / Volume of solution (in L)
- Density = Mass / Volume

Quick Tips:
- Always convert volume to liters when using molarity.
- Pay attention to units.

187 Medium

187 of 272

Application +4 | -1

If 65.5 g of a metal chloride contains 35.5 g chlorine, then equivalent mass of metal is

View answer and explanation

Correct option: A

30

Option A (30) is correct because the equivalent mass of the metal is calculated using the ratio of masses and equivalent masses of the metal and chlorine.

Solution approach

Correct Answer + Core Reason: 30. The equivalent mass is calculated using the mass ratio of metal and chlorine.

Step-by-Step Solution:

Step 1: Identify the given information: Mass of metal chloride = 65.5 g, Mass of chlorine = 35.5 g.

Step 2: Calculate the mass of the metal: Mass of metal = Mass of metal chloride - Mass of chlorine = 65.5 g - 35.5 g = 30 g.

Step 3: Apply the equivalent mass formula: $\frac{{W_{metal}}}{{E_{metal}}} = \frac{{W_{Cl}}}{{E_{Cl}}}$ , where W represents mass and E represents equivalent mass.

Step 4: Substitute the values: $\frac{{30}}{{E_{metal}}} = \frac{{35.5}}{{35.5}}$

Step 5: Solve for the equivalent mass of the metal: $E_{metal} = \frac{{30 \times 35.5}}{{35.5}} = 30$

Step 6: Final Answer: 30

Why other options are wrong:
- Option B (35.5): This is the mass of chlorine, not the equivalent mass of the metal.
- Option C (40.5): This value doesn't have a clear basis in the calculation and is likely a distractor.
- Option D (25): This value doesn't have a clear basis in the calculation and is likely a distractor.

Key Concepts:
- Equivalent mass
- Stoichiometry

Memory Aids:
- Equivalent mass formula: $\frac{{W_1}}{{E_1}} = \frac{{W_2}}{{E_2}}$

Quick Tips:
- Always ensure the masses used in the equivalent mass formula are for the elements combining directly.

188 Medium

188 of 272

Application +4 | -1

Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation,

2{\mathrm{H}_2}{\mathrm{O}_2}\left( {aq} \right) \to 2{\mathrm{H}_2}\mathrm{O}\left( l \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)

under conditions where one mole of gas occupies 24

\mathrm{d}{\mathrm{m}^3},100\mathrm{c}{\mathrm{m}^3}

of

XM

solution of

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

produces 3

\mathrm{d}{\mathrm{m}^3}

of

{\mathrm{O}_2}

. Thus,

X

is :

View answer and explanation

Correct option: A

2.5

Option A (2.5) is correct because the molarity X is calculated based on the volume of oxygen produced from the decomposition of hydrogen peroxide.

Solution approach

Correct Answer + Core Reason: 2.5. The molarity is calculated from the stoichiometry of the reaction and the volume of oxygen produced.

Step-by-Step Solution:

Step 1: Determine the moles of O2 produced.

Step 2: Moles of ${\mathrm{O}_2}$ formed $= \frac{3}{{24}} = \frac{1}{8}$ moles.

Step 3: Determine the moles of H2O2 required to produce that amount of O2. From the balanced equation, 2 moles of H2O2 produce 1 mole of O2.

Step 4: Therefore, moles of ${\mathrm{H}_2}{\mathrm{O}_2} = \frac{1}{8} \times 2 = \frac{1}{4}$ moles.

Step 5: Use the molarity formula to find X. Molarity (M) = moles / volume (in L). The volume is given as 100 cm³, which is 0.1 L.

Step 6: $X = \frac{{\frac{1}{4}}}{{0.1}} = \frac{1}{4} \times 10 = 2.5$ M

Why other options are wrong:
- Option B (1): This is incorrect as it doesn't account for the stoichiometry of the reaction and the conversion of volume.
- Option C (0.5): This is incorrect as it might be calculated by directly dividing the moles of oxygen by the volume without considering the stoichiometry.
- Option D (0.25): This is incorrect as it might be a result of dividing the moles of oxygen by the volume and also dividing by 2, which is not correct.

Key Concepts:
- Stoichiometry
- Molarity calculation
- Decomposition reaction

Memory Aids:
- Molarity = moles/volume

Quick Tips:
- Always balance the chemical equation first.
- Convert all volumes to the same units (usually liters).

189 Medium

189 of 272

Application +4 | -1

A transition metal

M

forms a volatile chloride which has a vapour density of 94.8 . If it contains

74.75 \%

of chlorine the formula of the metal chloride will be

View answer and explanation

Correct option: C

{MCl}_4

Option C (

{MCl}_4

) is correct because the valency of the metal is calculated to be 4, leading to the formula

{MCl}_4

.

Solution approach

Correct Answer + Core Reason: ${MCl}_4$ . The formula is determined by calculating the valency of the metal.

Step-by-Step Solution:

Step 1: Determine the mass of chlorine and the metal in 100 g of the metal chloride.\nMass of chlorine = 74.75 g\nMass of metal = 100 g - 74.75 g = 25.25 g

Step 2: Calculate the equivalent weight of the metal.\nEquivalent weight of metal = $\frac{{\text{{weight of metal}}}}{{\text{{weight of chlorine}}}} \times 35.5 = \frac{{25.25}}{{74.75}} \times 35.5 = 12$

Step 3: Calculate the valency of the metal using the vapor density.\nValency of metal = $\frac{{2 \times \text{{V.D.}}}}{{\text{{Equivalent wt. of metal}} + 35.5}} = \frac{{2 \times 94.8}}{{12 + 35.5}} = \frac{{189.6}}{{47.5}} = 4$

Step 4: Determine the formula of the metal chloride.\nSince the valency of the metal is 4, the formula of the metal chloride is ${MCl}_4$ .

Why other options are wrong:
- **Option A ( ${MCl}_3$ ): This is incorrect because the valency of the metal is not 3.
- Option B ( ${MCl}_2$ ): This is incorrect because the valency of the metal is not 2.
- Option D ( ${MCl}_5$ ): This is incorrect because the valency of the metal is not 5.

Key Concepts:
- Equivalent weight
- Valency
- Vapor density

Memory Aids:
- Remember the formula for calculating the valency of a metal chloride using vapor density.

Quick Tips:**
- Always start by finding the mass of each element in the compound.

190 Medium

190 of 272

Application +4 | -1

50 mL of an aqueous solution of glucose contains

6.02 \times {10^{22}}

molecules. The concentration of solution is:

View answer and explanation

Correct option: D

2.0 M

Option D (2.0 M) is correct because it is the concentration calculated using the number of molecules and volume of the solution.

Solution approach

Correct Answer + Core Reason: 2.0 M. The concentration of the glucose solution is calculated using the number of molecules and the volume of the solution.

Step-by-Step Solution:

Step 1: Calculate the number of moles of glucose: $\text{moles} = \frac{6.02 \times 10^{22}}{6.02 \times 10^{23}} = 0.1 \text{ moles}$ .

Step 2: Convert the volume of the solution to liters: $50 \text{ mL} = 0.05 \text{ L}$ .

Step 3: Calculate the molarity of the solution: $M = \frac{\text{moles}}{\text{volume}} = \frac{0.1 \text{ moles}}{0.05 \text{ L}} = 2.0 \text{ M}$ .

Final Answer: 2.0 M

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

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

Memory Aids:
- Remember the definition of molarity: moles of solute per liter of solution.

Quick Tips:
- Always convert the volume to liters before calculating molarity.
- Make sure to use Avogadro's number to convert between number of molecules and moles.

191 Medium

191 of 272

Application +4 | -1

Mass of one molecule of glucose is

\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}\right)

View answer and explanation

Correct option: D

3 \times 10^{-22} \mathrm{~g}

Option D (

3 \times 10^{-22} \mathrm{~g}

) is correct because it is calculated by dividing the molar mass of glucose by Avogadro's number.

Solution approach

Correct Answer + Core Reason: $3 \times 10^{-22} \mathrm{~g}$ . The mass of one molecule is calculated by dividing the molar mass by Avogadro's number.

Step-by-Step Solution:

Step 1: Identify the given information:\nMolar mass of glucose ( $C_6H_{12}O_6$ ) = 180 g/mol\nAvogadro's number ( $N_A$ ) = $6 \times 10^{23}$ molecules/mol

Step 2: Use the formula to calculate the mass of one molecule:\nMass of one molecule = $\frac{\text {Molar mass}}{N_A}$

Step 3: Substitute the values:\nMass of one molecule = $\frac{180 \mathrm{~g/mol}}{6 \times 10^{23} \mathrm{~molecules/mol}}$

Step 4: Calculate the mass:\nMass of one molecule = $3 \times 10^{-22} \mathrm{~g}$

Why other options are wrong:
- **Option A ( $3 \times 10^{-23} \mathrm{~g}$ ): This is incorrect because it is off by a factor of 10. It might be a calculation error.
- Option B ( $6 \times 10^{-24} \mathrm{~g}$ ): This is incorrect because it uses Avogadro's number in the denominator but doesn't divide the molar mass correctly.
- Option C ( $6 \times 10^{-23} \mathrm{~g}$ ): This is incorrect because it doesn't divide the molar mass by Avogadro's number correctly.

Key Concepts:
- Molar Mass
- Avogadro's Number

Memory Aids:**
- 1 mole = $N_A$ particles

Quick Tips:
- Always remember the units when performing calculations.
- Make sure to use the correct value for Avogadro's number.

192 Medium

192 of 272

Analyzing +4 | -1

For the reaction :

\mathrm{H}_3 \mathrm{PO}_3+\mathrm{Ba}(\mathrm{OH})_2 \rightarrow \mathrm{BaHPO}_3+2 \mathrm{H}_2 \mathrm{O}

The correct statement(s) is/are

(a) The equivalent mass of

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

is 41.

(b)

\mathrm{BaHPO}_3

is the acid salt

(c)

\mathrm{BaHPO}_3

is normal salt

(d) 1 mole of

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

is completely neutralized by 1.5 moles of

\mathrm{Ba}(\mathrm{OH})_2

View answer and explanation

Correct option: D

a and c are correct

Options a and c are correct because the equivalent mass of

H_3PO_3

is 41, and

BaHPO_3

is a normal salt.

Solution approach

Correct Answer + Core Reason: a and c are correct. The equivalent mass is 41 and $BaHPO_3$ is a normal salt.

Step-by-Step Solution:

Step 1: Analyze the reaction: $\mathrm{H}_3 \mathrm{PO}_3+\mathrm{Ba}(\mathrm{OH})_2 \rightarrow \mathrm{BaHPO}_3+2 \mathrm{H}_2 \mathrm{O}$

Step 2: $\mathrm{H}_3 \mathrm{PO}_3$ is a dibasic acid. It has two replaceable H atoms.

Step 3: Equivalent mass = Molar mass / Basicity = 82 / 2 = 41.

Step 4: $\mathrm{BaHPO}_3$ is a normal salt because all replaceable H atoms have been replaced by the metal.

Step 5: 1 mole of $\mathrm{H}_3 \mathrm{PO}_3$ is completely neutralized by 1 mole of $\mathrm{Ba}(\mathrm{OH})_2$ because it is dibasic.

Why other options are wrong:
- Option B: 'b' is incorrect because $BaHPO_3$ is a normal salt, not an acid salt.
- Option C: 'd' is incorrect because 1 mole of $\mathrm{H}_3 \mathrm{PO}_3$ is neutralized by 1 mole of $\mathrm{Ba}(\mathrm{OH})_2$ .
- Option D: 'b' is incorrect because $BaHPO_3$ is a normal salt, not an acid salt.

Key Concepts:
- Basicity of acids
- Equivalent mass
- Acid salts and normal salts

Memory Aids:
- Dibasic acids have two replaceable hydrogen atoms.

Quick Tips:
- Determine the basicity of the acid before calculating the equivalent mass.

193 Medium

193 of 272

Application +4 | -1

100 \mathrm{~mL}

of

\mathrm{N} / 5 \mathrm{NaOH}

will neutralize

View answer and explanation

Correct option: C

1.2368 \mathrm{~g}

of

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

Option C (

1.2368 \text{{ g}}

of

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

) is correct because it is calculated by equating the equivalents of NaOH and H3BO3, considering the n-factor of H3BO3 is 1.

Solution approach

Correct Answer + Core Reason: $1.2368 \text{{ g}}$ of $\mathrm{H}_3 \mathrm{BO}_3$ . The mass is calculated by equating equivalents of NaOH and H3BO3.

Step-by-Step Solution:

Step 1: Determine the n-factor of $\mathrm{H}_3 \mathrm{BO}_3$ . Boric acid is a monobasic acid, so its n-factor is 1.

Step 2: Calculate the equivalents of NaOH: Equivalents $= N \times V = \frac{1}{5} \times \frac{100}{1000} = 0.02 \text{{ equivalents}}$

Step 3: Equate the equivalents of $\mathrm{H}_3 \mathrm{BO}_3$ and NaOH: $\text{{Equivalents of }} \mathrm{H}_3 \mathrm{BO}_3 = \text{{Equivalents of NaOH}}$

Step 4: Calculate the mass of $\mathrm{H}_3 \mathrm{BO}_3$ : $\frac{x}{61.83} \times 1 = 0.02$ , where x is the mass of $\mathrm{H}_3 \mathrm{BO}_3$ and 61.83 is the molar mass of HBO3
$x = 0.02 \times 61.83 = 1.2366 \text{{ g}} \approx 1.2368 \text{{ g}}$

Why other options are wrong:
- **Option A ( $0.0618 \text{{ g}}$ of $\mathrm{H}_3 \mathrm{BO}_3$ ): This is incorrect because it likely involves an error in the calculation of equivalents or an incorrect molar mass.
- Option B ( $0.1855 \text{{ g}}$ of $\mathrm{H}_3 \mathrm{BO}_3$ ):** This is incorrect and suggests an underestimation of the mass of $\mathrm{H}_3 \mathrm{BO}_3$ , possibly due to an incorrect calculation of equivalents.
- **Option D ( $0.03092 \text{{ g}}$ of $\mathrm{H}_3 \mathrm{BO}_3$ ):** This is incorrect as it indicates a significant underestimation of the mass of $\mathrm{H}_3 \mathrm{BO}_3$ , possibly due to a significant error in the calculation.

Key Concepts:
- Normality
- Equivalents
- Neutralization reaction
- n-factor

Memory Aids:
- Equivalents = Normality * Volume
- n-factor of an acid is the number of replaceable H+ ions.

Quick Tips:
- Remember to determine the n-factor of the acid or base before calculation.
- Ensure units are consistent before calculation.

194 Medium

194 of 272

Analyzing +4 | -1

25 mL of 0.50

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

solution is added to 50 mL of 0.20 M

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

in acidic solution. Which of the following statements is true?

View answer and explanation

Correct option: B

0.005 mole of

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

are left

The correct answer is B (0.005 mole of

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

are left) because the milliequivalents of

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

exceed those of

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

, leaving excess

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

.

Solution approach

Correct Answer + Core Reason: 0.005 mole of $\mathrm{KMn}{\mathrm{O}_4}$ are left. This is because the initial amount of $\mathrm{KMn}{\mathrm{O}_4}$ is more than what is required to react with ${\mathrm{H}_2}{\mathrm{O}_2}$ .

Step-by-Step Solution:

Step 1: Calculate the milliequivalents of ${\mathrm{H}_2}{\mathrm{O}_2}$ .
Meq. of ${\mathrm{H}_2}{\mathrm{O}_2} = 25 \times 0.50 \times 2 = 25$

Step 2: Calculate the milliequivalents of $\mathrm{KMn}{\mathrm{O}_4}$ .
Meq. of $\mathrm{KMn}{\mathrm{O}_4} = 50 \times 0.20 \times 5 = 50$

Step 3: Determine the limiting reactant. Since the milliequivalents of $\mathrm{KMn}{\mathrm{O}_4}$ (50) are greater than those of ${\mathrm{H}_2}{\mathrm{O}_2}$ (25), ${\mathrm{H}_2}{\mathrm{O}_2}$ is the limiting reactant.

Step 4: Calculate the milliequivalents of $\mathrm{KMn}{\mathrm{O}_4}$ left unreacted.
Meq. of $\mathrm{KMn}{\mathrm{O}_4}$ left = 50 - 25 = 25

Step 5: Convert milliequivalents of $\mathrm{KMn}{\mathrm{O}_4}$ left to moles.
Moles of $\mathrm{KMn}{\mathrm{O}_4}$ left = 25 meq / 5 (eq/mol) / 1000 (meq/eq) = 0.005 moles

Why other options are wrong:
- Option A (0.010 mole of oxygen is liberated): The amount of oxygen liberated is related to the amount of ${\mathrm{H}_2}{\mathrm{O}_2}$ reacted. This calculation is incorrect.
- Option C (0.030 g atom of oxygen gas is evolved): This is equivalent to 0.03 moles of oxygen atoms, which is not consistent with the amount of ${\mathrm{H}_2}{\mathrm{O}_2}$ reacted.
- **Option D (0.0025 mole ${\mathrm{H}_2}{\mathrm{O}_2}$ does not react with $\mathrm{KMn}{\mathrm{O}_4}$ ):** All of the ${\mathrm{H}_2}{\mathrm{O}_2}$ reacts because it is the limiting reactant.

Key Concepts:
- Redox reactions
- Stoichiometry
- Limiting reactant
- Milliequivalents

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

Quick Tips:
- Always calculate milliequivalents to determine the limiting reactant in redox titrations.

195 Medium

195 of 272

Application +4 | -1

When

10 \mathrm{~ml}

of acetic acid (density

=0.8 \mathrm{~g} / \mathrm{ml}

) is mixed with

40 \mathrm{~ml}

of water (density

=1 \mathrm{~g} / \mathrm{ml}

) at a certain temperature, the final solution is found to have a density of

\frac{96}{98} \mathrm{~g} / \mathrm{ml}.

The percent change in total volume on mixing is

View answer and explanation

Correct option: A

2 \%

Option A (

2 \%

) is correct because it is the percent change in volume calculated from the initial and final volumes after mixing.

Solution approach

Correct Answer + Core Reason: $2 \%$ . The percent change in volume is calculated by comparing the initial and final volumes of the solution.

Step-by-Step Solution:

Step 1: Calculate the mass of acetic acid:\nVolume of acetic acid = 10 mL\nDensity of acetic acid = 0.8 g/mL\nMass of acetic acid = Volume × Density = 10 mL × 0.8 g/mL = 8 g

Step 2: Calculate the mass of water:\nVolume of water = 40 mL\nDensity of water = 1 g/mL\nMass of water = Volume × Density = 40 mL × 1 g/mL = 40 g

Step 3: Calculate the total mass of the solution:\nTotal mass = Mass of acetic acid + Mass of water = 8 g + 40 g = 48 g

Step 4: Calculate the final volume of the solution:\nDensity of the final solution = $\frac{96}{98}$ g/mL\nFinal volume = $\frac{{\text{{Total mass}}}}{{\text{{Density}}}} = \frac{{48 \text{{ g}}}}{{\frac{{96}}{{98}} \text{{ g/mL}}}} = 48 \times \frac{{98}}{{96}} \text{{ mL}} = 49 \text{{ mL}}$

Step 5: Calculate the initial volume of the solution:\nInitial volume = Volume of acetic acid + Volume of water = 10 mL + 40 mL = 50 mL

Step 6: Calculate the percent change in volume:\nPercent change = $\frac{{\text{{Initial volume}} - \text{{Final volume}}}}{{\text{{Initial volume}}}} \times 100 = \frac{{50 - 49}}{{50}} \times 100 = \frac{1}{{50}} \times 100 = 2 \%$

Final Answer: $2 \%$

Why other options are wrong:
- **Option B ( $4 \%$ ): This is incorrect; it might be a calculation error.
- Option C ( $9 \%$ ): This is incorrect; it might be a calculation error.
- Option D ( $5 \%$ ): This is incorrect; it might be a calculation error.

Key Concepts:
- Density
- Volume
- Mass
- Percent change

Memory Aids:**
- Remember the formula: Density = $\frac{{\text{{Mass}}}}{{\text{{Volume}}}}$

Quick Tips:
- Ensure the units are consistent before performing calculations.
- Pay attention to the given densities and volumes.

196 Medium

196 of 272

Application +4 | -1

At NTP 2.24L of a gas weight

3.65 \mathrm{~g}

. The mole of the gas in

18.25 \mathrm{~g}

is :-

View answer and explanation

Correct option: B

0.5

Option B (0.5) is correct because it is the number of moles calculated using the given volume and mass of the gas at NTP.

Solution approach

Correct Answer + Core Reason: 0.5. The number of moles is calculated using the molar mass derived from the given volume and mass at NTP.

Step-by-Step Solution:

Step 1: At NTP, 2.24 L of the gas weighs 3.65 g.

Step 2: Calculate the mass of 22.4 L (1 mole) of the gas: Mass of 1 mole = $\frac{3.65 \text{{ g}}}{2.24 \text{{ L}}} \times 22.4 \text{{ L}} = 36.5 \text{{ g}}$ .

Step 3: Calculate the number of moles in 18.25 g of the gas: Moles = $\frac{18.25 \text{{ g}}}{36.5 \text{{ g/mol}}} = 0.5 \text{{ mol}}$ .

Step 4: Final Answer: The number of moles of the gas in 18.25 g is 0.5.

Why other options are wrong:
- Option A (0.25): This is incorrect due to a miscalculation in the mole conversion.
- Option C (0.75): This is incorrect due to a miscalculation in the mole conversion.
- Option D (1): This is incorrect because it assumes 36.5 g is 36.5 g/mol and calculates moles for 36.5g.

Key Concepts:
- Mole concept
- NTP conditions

Memory Aids:
- At NTP, 1 mole of any gas occupies 22.4 L.

Quick Tips:
- Use the molar volume at NTP to find the molar mass of the gas.
- Convert mass to moles using the molar mass.

197 Medium

197 of 272

Analyzing +4 | -1

Assertion (A) : In the following half reaction:

\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow \mathrm{Cr}^{3+}

Equivalent weight of

\mathrm{Cr}_2 \mathrm{O}_7^{2-}=\frac{\text { Molar mass }}{3}

Reason (R) : Change in oxidation number of chromium is 3 per atom.

View answer and explanation

Correct option: D

\mathbf{A}

is false but

\mathbf{R}

is true.

Option D is correct because the assertion is false; the equivalent weight is molar mass divided by 6, not 3. The reason is true; the change in oxidation number of chromium is 3 per atom.

Solution approach

Correct Answer + Core Reason: $\mathbf{A}$ is false but $\mathbf{R}$ is true. The equivalent weight calculation is incorrect, but the oxidation state change is correct.

Step-by-Step Solution:

Step 1: Determine the oxidation state change of chromium in the reaction $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow \mathrm{Cr}^{3+}$ .

Step 2: In $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ , the oxidation state of Cr is +6. In $\mathrm{Cr}^{3+}$ , the oxidation state of Cr is +3. The change in oxidation number per chromium atom is 3.

Step 3: Since there are two chromium atoms in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ , the total change in oxidation number per molecule is 2 * 3 = 6.

Step 4: The equivalent weight is calculated as $\frac{{\text{{Molar mass}}}}{{\text{{Total change in oxidation number}}}} = \frac{{\text{{Molar mass}}}}{{6}}$ .

Step 5: Therefore, the assertion is false, and the reason is true.

Why other options are wrong:
- Option A: This is incorrect because the assertion is false.
- Option B: This is incorrect because the assertion is false.
- Option C: This is incorrect because the reason is true.

Key Concepts:
- Oxidation state
- Equivalent weight
- Redox reactions

Memory Aids:
- Oxidation is Loss of electrons, Reduction is Gain of electrons (OIL RIG)
- Equivalent weight = Molar mass / n-factor

Quick Tips:
- Always balance the redox reaction before determining the n-factor.
- Remember to consider the number of atoms involved when calculating the total change in oxidation number.

198 Medium

198 of 272

Application +4 | -1

What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene

View answer and explanation

Correct option: C

9.6 kg

Option C (9.6 kg) is correct because it is the calculated mass of oxygen required for the complete combustion of 2.8 kg of ethylene based on stoichiometry.

Solution approach

Correct Answer + Core Reason: 9.6 kg. The mass of oxygen is calculated based on the stoichiometry of the combustion reaction.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the combustion of ethylene:
$C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g)$

Step 2: Calculate the molar mass of ethylene ( $C_2H_4$ ):
2(12) + 4(1) = 24 + 4 = 28 g/mol

Step 3: Calculate the molar mass of oxygen ( $O_2$ ):
2(16) = 32 g/mol

Step 4: Calculate the number of moles of ethylene:\nMoles of $C_2H_4$ = $\frac{2.8 \text{ kg}}{28 \text{ g/mol}} = \frac{2800 \text{ g}}{28 \text{ g/mol}} = 100 \text{ moles}$

Step 5: From the balanced equation, 1 mole of $C_2H_4$ requires 3 moles of $O_2$ .\nSo, 100 moles of $C_2H_4$ require 300 moles of $O_2$ .

Step 6: Calculate the mass of oxygen required:\nMass of $O_2$ = 300 moles * 32 g/mol = 9600 g = 9.6 kg

Step 7: Final Answer: 9.6 kg

Why other options are wrong:
- Option A: 2.8 kg is incorrect because it is the mass of ethylene, not oxygen.
- Option B: 6.4 kg is incorrect and likely arises from an incorrect calculation of the stoichiometry.
- Option D: 96 kg is incorrect and likely arises from a miscalculation.

Key Concepts:
- Stoichiometry
- Combustion Reactions
- Molar Mass

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

Quick Tips:
- Always convert mass to moles before using stoichiometry.
- Pay attention to the units and make sure they are consistent.

199 Medium

199 of 272

Analyzing +4 | -1

In the following reaction,

\mathrm{As}_2 \mathrm{~S}_3+\mathrm{H}^{+}+\mathrm{NO}_3^{-} \longrightarrow \mathrm{NO}+\mathrm{H}_2 \mathrm{O}+\mathrm{AsO}_4^{3-}+\mathrm{SO}_4^{2-}

the equivalent weight of

\mathrm{As}_2 \mathrm{~S}_3

is related to its molecular weight by :

View answer and explanation

Correct option: D

M / 28

Option D is correct because the n-factor is calculated based on the change in oxidation states of As and S, leading to M/28.

Solution approach

Correct Answer + Core Reason: $M / 28$ . The equivalent weight is calculated by dividing the molecular weight by the n-factor, which is 28.

Step-by-Step Solution:
Step 1: Write the balanced chemical equation:
$\mathrm{As}_2 \mathrm{~S}_3+\mathrm{H}^{+}+\mathrm{NO}_3^{-} \longrightarrow \mathrm{NO}+\mathrm{H}_2 \mathrm{O}+\mathrm{AsO}_4^{3-}+\mathrm{SO}_4^{2-}$
Step 2: Determine the change in oxidation state for As and S.
$2 \mathrm{As}^{+3} \rightarrow 2 \mathrm{As}^{+5}$ (change = $2 \times 2 = 4$ )
$3 \mathrm{~S}^{-2} \rightarrow 3 \mathrm{~S}^{+6}$ (change = $3 \times 8 = 24$ )
Step 3: Calculate the n-factor: n-factor $= |2 \times 2+8 \times 3| = |4 + 24| = 28$
Step 4: Equivalent weight $= \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{M}{28}$

Why other options are wrong:
- Option A: Incorrect n-factor.
- Option B: Incorrect n-factor.
- Option C: Incorrect n-factor.

Key Concepts:
- Redox Reactions
- Equivalent Weight
- n-factor Calculation

Memory Aids:
- Remember that n-factor is the total change in oxidation state.

Quick Tips:
- Balance the equation first to correctly determine the change in oxidation states.

200 Medium

200 of 272

Analyzing +4 | -1

Assertion (A) : 1 mole of

\mathrm{FeC}_2 \mathrm{O}_4

is oxidised by 0.6 mole of

\mathrm{MnO}_4^{-}

in acidic medium.

Reason (R) :

\mathrm{MnO}_4^{-}

oxidises both

\mathrm{Fe}^{2+}

as well as

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

.

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}

.

Option A is correct because both the assertion and the reason are true, and the reason correctly explains why 1 mole of

\mathrm{FeC}_2 \mathrm{O}_4

requires 0.6 moles of

\mathrm{MnO}_4^{-}

for oxidation in acidic medium.

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 amount of $\mathrm{MnO}_4^-$ required is determined by the oxidation of both $\mathrm{Fe}^{2+}$ and $\mathrm{C}_2\mathrm{O}_4^{2-}$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equations for the oxidation of $\mathrm{Fe}^{2+}$ and $\mathrm{C}_2\mathrm{O}_4^{2-}$ by $\mathrm{MnO}_4^-$ in acidic medium.
$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-$ (n-factor = 1)
$\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{CO}_2 + 2e^-$ (n-factor = 2)
$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$ (n-factor = 5)

Step 2: Calculate the total n-factor for $\mathrm{FeC}_2\mathrm{O}_4$ .\nThe n-factor for $\mathrm{FeC}_2\mathrm{O}_4$ is the sum of the n-factors for $\mathrm{Fe}^{2+}$ and $\mathrm{C}_2\mathrm{O}_4^{2-}$ , which is 1 + 2 = 3.

Step 3: Use the n-factor concept to find the moles of $\mathrm{MnO}_4^-$ required.
$n_1 \times moles_1 = n_2 \times moles_2$
$3 \times 1 = 5 \times moles(\mathrm{MnO}_4^-)$ \nmoles of $\mathrm{MnO}_4^- = \frac{3}{5} = 0.6$

Step 4: Conclusion\nTherefore, 1 mole of $\mathrm{FeC}_2\mathrm{O}_4$ is oxidized by 0.6 moles of $\mathrm{MnO}_4^-$ .

Why other options are wrong:
- Option B: While both A and R are true, R is the correct explanation, making B incorrect.
- Option C: A is true, so C is incorrect.
- Option D: A is true, so D is incorrect.

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

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

Quick Tips:
- Balance the half-reactions before calculating n-factors.
- Remember to consider the number of atoms when calculating n-factors for molecules like $\mathrm{C}_2\mathrm{O}_4^{2-}$ .
- The n-factor is the number of electrons transferred in a redox reaction.

201 Medium

201 of 272

Application +4 | -1

The element

\mathrm{Q}

has the following isotopic composition,

\mathrm{Q}^{200}: 80 \%; \mathrm{Q}^{190}: 15 \%

and

\mathrm{Q}^{210}: 5 \%

. The average atomic mass of the naturally occurring element

\mathrm{Q}

is

View answer and explanation

Correct option: C

199 \mathrm{~u}

Option C (199 u) is correct because it's calculated as the weighted average of the isotopic masses based on their abundances.

Solution approach

Correct Answer + Core Reason: 199 u. The average atomic mass is calculated as the weighted average of isotopic masses.

Step-by-Step Solution:

Step 1: Average atomic mass = (Atomic mass of $Q^{200} \times$ abundance + atomic mass of $Q^{190} \times$ abundance + atomic mass of $Q^{210} \times$ abundance) / 100

Step 2: Average atomic mass = $\frac{{200 \times 80 + 190 \times 15 + 210 \times 5}}{{100}}$

Step 3: Average atomic mass = $\frac{{16000 + 2850 + 1050}}{{100}} = 199 u$

Why other options are wrong:
- Option A (200 u): This is incorrect as it only considers the most abundant isotope and ignores the contributions of the other isotopes.
- Option B (198 u): This value is incorrect due to a miscalculation in the weighted average.
- Option D (202 u): This value is incorrect due to a miscalculation in the weighted average.

Key Concepts:
- Average atomic mass
- Isotopic abundance

Memory Aids:
- Weighted average formula: $\sum (mass \times abundance) / 100$

Quick Tips:
- Remember to divide by 100 after summing the weighted masses.

202 Medium

202 of 272

Remembering +4 | -1

The product of atomic weight and specific heat of a metal is approximately 6.4. This was given by:

View answer and explanation

Correct option: D

DulongPetit’s law

Option D (DulongPetit’s law) is correct because Dulong-Petit's law states that the product of atomic weight and specific heat of a metal is approximately 6.4.

Solution approach

Correct Answer + Core Reason: DulongPetit’s law. This law relates atomic weight and specific heat.

Step-by-Step Solution:

Step 1: Recall Dulong-Petit's Law.

Step 2: Dulong-Petit's Law states: Atomic weight * Specific heat ≈ 6.4 cal/mol·K

Step 3: Identify the correct law.

Step 4: The question directly asks for the law that relates atomic weight and specific heat to approximately 6.4.

Step 5: Dulong-Petit's Law is the correct answer.

Why other options are wrong:
- Option A: Dalton's law relates to partial pressures of gases in a mixture, not specific heat.
- Option B: Avogadro's law relates to the volume and number of moles of a gas, not specific heat.
- Option C: Newton's law describes motion and gravity, not specific heat.

Key Concepts:
- Dulong-Petit's Law

Memory Aids:
- Remember Dulong-Petit for specific heat capacity of solids.

Quick Tips:
- This is a direct recall question. Knowing the laws is essential.

203 Medium

203 of 272

Application +4 | -1

In a rocket motor filled with butane, how many

\mathrm{kg}

of liquid oxygen should be provided for each

\mathrm{kg}

of butane for its complete combustion?

\mathrm{C}_4 \mathrm{H}_{10}+13 / 2 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}

View answer and explanation

Correct option: A

4 kg

The correct answer is 4 kg because it represents the amount of liquid oxygen needed for complete combustion of 1 kg of butane based on the stoichiometry of the balanced chemical equation.

Solution approach

Correct Answer + Core Reason: 4 kg. The amount of liquid oxygen required is calculated using stoichiometry.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the combustion of butane.
$C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$

Step 2: Determine the molar masses of butane and oxygen.\nMolar mass of butane ( $C_4H_{10}$ ) = (4 × 12) + (10 × 1) = 48 + 10 = 58 g/mol\nMolar mass of oxygen ( $O_2$ ) = 2 × 16 = 32 g/mol

Step 3: Calculate the mass of oxygen required to react with 1 kg of butane.\nFrom the balanced equation, 1 mole of butane reacts with $\frac{13}{2}$ moles of oxygen.\nMass of oxygen = $\frac{\text{Mass of butane (in grams)}}{\text{Molar mass of butane}} \times \frac{\text{Moles of oxygen}}{\text{Moles of butane}} \times \text{Molar mass of oxygen}$ \nMass of oxygen = $\frac{1000}{58} \times \frac{13}{2} \times 32 = 3586.2 \text{ g} \approx 3.586 \text{ kg}$

Step 4: Round the answer to the nearest whole number.\nMass of oxygen ≈ 4 kg

Why other options are wrong:
- Option A: Incorrect stoichiometric calculation.
- Option B: Incorrect stoichiometric calculation.
- Option C: Incorrect stoichiometric calculation.
- Option D: Incorrect stoichiometric calculation.

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

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

Quick Tips:
- Always convert masses to moles before using stoichiometric ratios.

204 Medium

204 of 272

Application +4 | -1

0.20 g of a hydrocarbon on combustion gave 0.66 g

C{O_2}.

The percentage of hydrogen in the hydrocarbon is about:

View answer and explanation

Correct option: C

10

Option C (10) is correct because the percentage of hydrogen is calculated by subtracting the percentage of carbon from 100.

Solution approach

Correct Answer + Core Reason: 10. The percentage of hydrogen is determined by calculating the percentage of carbon and subtracting it from 100.

Step-by-Step Solution:

Step 1: Calculate the mass of carbon in 0.66 g of $CO_2$ . The molar mass of $CO_2$ is 44 g/mol, and the molar mass of carbon is 12 g/mol.

Step 2: The mass of carbon is $\frac{{12}}{{44}} \times 0.66 \text{{ g}} = 0.18 \text{{ g}}$ .

Step 3: Calculate the percentage of carbon in the hydrocarbon: $\% \,C = \frac{{0.18}}{{0.20}} \times 100 = 90\%$ .

Step 4: Calculate the percentage of hydrogen in the hydrocarbon: $\% \,H = 100\% - 90\% = 10\%$ .

Final Answer: The percentage of hydrogen in the hydrocarbon is 10%.

Why other options are wrong:
- Option A: This value (33) is incorrect. It might be a calculation error.
- Option B: This value (45) is incorrect. It might be a calculation error.
- Option D: This value (90) is incorrect. This is the percentage of carbon, not hydrogen.

Key Concepts:
- Combustion analysis
- Percentage composition
- Stoichiometry

Memory Aids:
- Remember that the percentage of elements in a compound must add up to 100%.

Quick Tips:
- Always check the molar masses of the compounds involved in the calculation.
- Ensure the units are consistent.

205 Medium

205 of 272

Application +4 | -1

A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of

C{O_2}.

The empirical formula of the hydrocarbon is

View answer and explanation

Correct option: D

{C_7}{H_8}

Option D (

{C_7}{H_8}

) is correct because the ratio of moles of Carbon to Hydrogen is 7:8, which is determined from the combustion products.

Solution approach

Correct Answer + Core Reason: ${C_7}{H_8}$ . The empirical formula is determined by calculating the mole ratio of carbon to hydrogen.

Step-by-Step Solution:

Step 1: Calculate the mass of hydrogen in 0.72 g of water (H2O).\nMolar mass of H2O = 2(1) + 16 = 18 g/mol\nMass of hydrogen in 1 mole of H2O = 2 g\nSo, 18 g of H2O contains 2 g of H\nTherefore, 0.72 g of H2O contains (2/18) * 0.72 = 0.08 g of H

Step 2: Calculate the mass of carbon in 3.08 g of carbon dioxide (CO2).\nMolar mass of CO2 = 12 + 2(16) = 44 g/mol\nMass of carbon in 1 mole of CO2 = 12 g\nSo, 44 g of CO2 contains 12 g of C\nTherefore, 3.08 g of CO2 contains (12/44) * 3.08 = 0.84 g of C

Step 3: Calculate the moles of carbon and hydrogen.\nMoles of C = mass / molar mass = 0.84 g / 12 g/mol = 0.07 mol\nMoles of H = mass / molar mass = 0.08 g / 1 g/mol = 0.08 mol

Step 4: Determine the mole ratio of C to H.\nC : H = 0.07 : 0.08\nDivide both by the smallest number (0.07) to simplify the ratio:\nC : H = 1 : (0.08/0.07) = 1 : 1.14 ≈ 1 : 8/7\nMultiply both sides by 7 to get whole numbers:\nC : H = 7 : 8

Step 5: Write the empirical formula.\nThe empirical formula is ${C_7}{H_8}$

Why other options are wrong:
- Option A: The mole ratio of C to H is not 2:4.
- Option B: The mole ratio of C to H is not 3:4.
- Option C: The mole ratio of C to H is not 6:5.

Key Concepts:
- Empirical formula
- Combustion analysis
- Mole concept

Memory Aids:
- Remember to convert masses to moles before determining the ratio.

Quick Tips:
- Always simplify the mole ratio to the smallest whole numbers.

206 Medium

206 of 272

Application +4 | -1

7.5 \mathrm{~gs}

of a gas occupies

5.8 \mathrm{~L}

of volume at STP. The gas is:

View answer and explanation

Correct option: A

\mathrm{NO}

Option A (

\mathrm{NO}

) is correct because the molar mass calculated from the given volume and mass at STP is approximately 30 g/mol, which corresponds to nitric oxide (NO).

Solution approach

Correct Answer + Core Reason: $\mathrm{NO}$ . The gas is identified by calculating its molar mass.

Step-by-Step Solution:

Step 1: Identify the given information: Mass of gas = 7.5 g, Volume of gas = 5.8 L at STP.

Step 2: Calculate the volume occupied by 1 mole of gas at STP: 22.4 L.

Step 3: Calculate the molar mass of the gas: If 5.8 L has a mass of 7.5 g, then 22.4 L has a mass of $\frac{7.5}{5.8} \times 22.4 = 28.96 \text{ g/mol}$ .

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 calculated molar mass (28.96 g/mol) is closest to NO (30 g/mol).

Step 6: Final Answer: The gas is NO.

Why other options are wrong:
- **Option B ( $N_2O$ ):** The molar mass of $N_2O$ is 44 g/mol, which is significantly different from the calculated molar mass.
- Option C (CO): The molar mass of CO is 28 g/mol, which is close but less accurate than NO.
- **Option D ( $CO_2$ ):** The molar mass of $CO_2$ is 44 g/mol, which is significantly different from the calculated molar mass.

Key Concepts:
- Molar mass
- STP conditions
- Ideal gas law

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

Quick Tips:
- Calculate the molar mass using the given mass and volume at STP.
- Compare the calculated molar mass with the molar masses of the given options.

207 Medium

207 of 272

Application +4 | -1

Versene, a chelating agent having chemical formula

{\mathrm{C}_2}{\mathrm{H}_4}{\mathrm{N}_2}{\left( {{\mathrm{C}_2}{\mathrm{H}_2}{\mathrm{O}_2}\mathrm{Na}} \right)_4}

. If each mole of this compound could bind 1 mole of

\mathrm{C}{\mathrm{a}^{2 + }}

,then the rating of pure versene expressed as mg of

\mathrm{CaC}{\mathrm{O}_3}

bound per g of chelating agent is:

View answer and explanation

Correct option: D

263 mg

The correct answer is D (263 mg) because the rating of pure versene is calculated based on the molar mass of CaCO3 and the chelating agent.

Solution approach

Correct Answer + Core Reason: 263 mg. The rating is calculated based on the molar mass of CaCO3 and the chelating agent.

Step-by-Step Solution:

Step 1: Identify the given information: 1 mole of Versene binds 1 mole of $\mathrm{Ca}^{2+}$ . The molar mass of $\mathrm{CaCO}_3$ is 100 g/mol.

Step 2: Determine the molar mass of Versene: ${\mathrm{C}_2}{\mathrm{H}_4}{\mathrm{N}_2}{\left( {{\mathrm{C}_2}{\mathrm{H}_2}{\mathrm{O}_2}\mathrm{Na}} \right)_4}$ . The molar mass is approximately 380 g/mol (given in the solution).

Step 3: Calculate the rating of pure Versene as mg of $\mathrm{CaCO}_3$ bound per g of chelating agent.

Step 4: 1 mole $\mathrm{Ca}^{2 + } = 1 \;\mathrm{mole}\;\mathrm{CaCO}_3 = 100\;\mathrm{g} = 100 \times 10^3 \;\mathrm{mg}$ .

Step 5: Rating $= \frac{{\text{mg of }{\mathrm{CaCO}_3}}}{{\text{g of chelating agent}}} = \frac{{100 \times {{10}^3}}}{{380}} \approx 263 \;\mathrm{mg}$ .

Final Answer: 263 mg

Why other options are wrong:
- Option A (100 mg): This is incorrect because it doesn't take into account the molar mass of the chelating agent.
- Option B (163 mg): This is incorrect due to a calculation error.
- Option C (200 mg): This is incorrect due to a calculation error.

Key Concepts:
- Chelating agents
- Molar mass calculations
- Stoichiometry

Memory Aids:
- Remember to convert grams to milligrams when necessary.

Quick Tips:
- Always check the units and make sure they are consistent throughout the calculation.

208 Medium

208 of 272

Application +4 | -1

A solution is prepared by dissolving

24.5 {~g}

of sodium hydroxide in distilled water to give

1 {~L}

solution. The molarity of

{NaOH}

in the solution is
(Given that molar mass of

{NaOH}=40.0 {~g} {~mol}^{-1}

)

View answer and explanation

Correct option: B

0.6125 {M}

Option B (

0.6125 {M}

) is correct because it is the molarity of NaOH calculated using the given mass and volume.

Solution approach

Correct Answer + Core Reason: $0.6125 {M}$ . The molarity of NaOH is calculated using the formula: Molarity = moles/volume.

Step-by-Step Solution:

Step 1: Identify the given information: Mass of NaOH = 24.5 g, Volume of solution = 1 L, Molar mass of NaOH = 40.0 g/mol.

Step 2: Calculate the number of moles of NaOH: Moles = $\frac{{Mass}}{{Molar Mass}} = \frac{{24.5 g}}{{40.0 g/mol}} = 0.6125 mol$ .

Step 3: Calculate the molarity of the solution: Molarity = $\frac{{Moles}}{{Volume}} = \frac{{0.6125 mol}}{{1 L}} = 0.6125 M$ .

Step 4: Final Answer: $0.6125 {M}$

Why other options are wrong:
- **Option A ( $0.2450 {M}$ ): This is incorrect because it likely results from dividing the mass by 100 instead of the molar mass.
- Option C ( $0.9800 {M}$ ): This is incorrect because it likely results from a calculation error.
- Option D ( $1.6326 {M}$ ): This is incorrect because it likely results from a calculation error.

Key Concepts:
- Molarity
- Moles

Memory Aids:
- Molarity = Moles of solute / Liters of solution.

Quick Tips:**
- Ensure the units are consistent.
- Use the correct molar mass.

209 Medium

209 of 272

Application +4 | -1

Weight of

\mathrm{FeS}{\mathrm{O}_4}\left( {\mathrm{mol}\mathrm{.wt}\mathrm{.} = 152} \right)

oxidized by 200 mL of 1

N

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

solution is :

View answer and explanation

Correct option: A

30.4 g

Option A (30.4 g) is correct because it is calculated using the concept of gram equivalents and the balanced redox reaction.

Solution approach

Correct Answer + Core Reason: 30.4 g. The weight of FeSO4 oxidized is calculated using the concept of equivalents.

Step-by-Step Solution:

Step 1: Understand the concept of equivalents: Equivalents of $FeSO_4$ = Equivalents of $KMnO_4$

Step 2: Calculate the milliequivalents of $KMnO_4$ : Milliequivalents = Normality x Volume (in mL) = 1 N x 200 mL = 200 meq

Step 3: Determine the equivalent weight of $FeSO_4$ . In this reaction, $Fe^{2+}$ is oxidized to $Fe^{3+}$ , so the n-factor is 1. Therefore, Equivalent weight = Molecular weight / n-factor = 152 / 1 = 152

Step 4: Calculate the weight of $FeSO_4$ : Weight = (Milliequivalents x Equivalent weight) / 1000 = (200 x 152) / 1000 = 30.4 g

Why other options are wrong:
- Option B (15.2 g): This is incorrect because it likely involves dividing the equivalent weight by 2 instead of using the correct n-factor of 1.
- Option C (60.8 g): This is incorrect because it likely involves multiplying the equivalent weight by 2 instead of using the correct n-factor of 1.
- Option D (158 g): This is incorrect and likely involves a misunderstanding of the reaction stoichiometry or incorrect use of the formula.

Key Concepts:
- Redox reactions
- Equivalent weight
- Normality
- Gram equivalents

Memory Aids:
- Remember that equivalents of reactants are equal in a balanced redox reaction.

Quick Tips:
- Always determine the n-factor (number of electrons transferred) correctly.
- Use the formula: Weight = (Equivalents x Equivalent weight) / 1000

210 Medium

210 of 272

Application +4 | -1

1 gram of sodium hydroxide was treated with

25 \mathrm{~mL}

of

0.75 \mathrm{~M} \, {HCl}

solution, the mass of sodium hydroxide left unreacted is equal to

View answer and explanation

Correct option: A

250 \mathrm{~mg}

Option A (250 mg) is correct because after reacting 1 g of NaOH with 25 mL of 0.75 M HCl, 250 mg of NaOH remains unreacted.

Solution approach

Correct Answer + Core Reason: 250 mg. The mass of unreacted NaOH is calculated by finding the difference between the initial mass and the mass that reacted with HCl.

Step-by-Step Solution:

Step 1: Calculate the moles of HCl used: Molarity (M) = Moles / Volume (L), so Moles = Molarity x Volume = 0.75 M x 0.025 L = 0.01875 moles.

Step 2: Write the balanced chemical equation: $HCl + NaOH \longrightarrow NaCl + H_2O$ . The mole ratio of HCl to NaOH is 1:1.

Step 3: Calculate the moles of NaOH that reacted with HCl: Since the mole ratio is 1:1, moles of NaOH reacted = moles of HCl = 0.01875 moles.

Step 4: Calculate the mass of NaOH that reacted: Mass = Moles x Molar mass = 0.01875 moles x 40 g/mol = 0.75 g.

Step 5: Calculate the initial mass of NaOH: 1 g.

Step 6: Calculate the mass of NaOH left unreacted: Mass unreacted = Initial mass - Mass reacted = 1 g - 0.75 g = 0.25 g = 250 mg.

Why other options are wrong:
- Option B (Zero mg): This would be true if all NaOH reacted, but it didn't.
- Option C (200 mg): This is a calculation error.
- Option D (750 mg): This is the mass of NaOH that reacted, not the mass left unreacted.

Key Concepts:
- Molarity
- Stoichiometry
- Mole concept

Memory Aids:
- Molarity = Moles / Volume
- Mass = Moles x Molar mass

Quick Tips:
- Always write the balanced chemical equation first.
- Pay attention to units.

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