SOME BASIC CONCEPTS OF CHEMISTRY MCQs for JEE CHEMISTRY — Medium

1 Medium

1 of 272

Application +4 | -1

One litre of

{CO}_2

is passed through red hot coke. The volume becomes 1.4 litres at same temperature and pressure. The composition of products is

View answer and explanation

Correct option: C

0.6 litre of

{CO}_2

and 0.8 litre of

{CO}

Option C (0.6 litre of

{CO}_2

and 0.8 litre of

{CO}

) is correct because the reaction consumes some CO2 to produce CO, and the final volume is 1.4L.

Solution approach

Correct Answer + Core Reason: 0.6 litre of ${CO}_2$ and 0.8 litre of ${CO}$ . The composition is determined by stoichiometry and the change in volume.

Step-by-Step Solution:
Step 1: Write the reaction: $\mathrm{CO_2(g) + C(s) \longrightarrow 2CO(g)}$
Step 2: Set up the initial and final volumes: Initial: 1 L $\mathrm{CO_2}$ , 0 L $\mathrm{CO}$ . Final: (1-x) L $\mathrm{CO_2}$ , 2x L $\mathrm{CO}$
Step 3: Calculate the final volume: $\text{Final volume} = (1-x) + 2x = 1 + x = 1.4 \text{ L}$
Step 4: Solve for x: $x = 0.4 \text{ L}$
Step 5: Calculate the final volumes of $\mathrm{CO_2}$ and $\mathrm{CO}$ : $\mathrm{CO_2} = 1 - x = 1 - 0.4 = 0.6 \text{ L}$ , $\mathrm{CO} = 2x = 2 \times 0.4 = 0.8 \text{ L}$
Step 6: Conclude: The composition is 0.6 L of $\mathrm{CO_2}$ and 0.8 L of $\mathrm{CO}$ .

Why other options are wrong:
- Option A: 0.8 litre of ${CO}_2$ and 0.6 litre of ${CO}$ is incorrect because the volumes are reversed.
- Option B: 0.7 litre of ${CO}_2$ and 0.7 litre of ${CO}$ is incorrect because it doesn't satisfy the stoichiometry and volume change.
- Option D: 0.4 litre of ${CO}_2$ and 1.0 litre of ${CO}$ is incorrect because it doesn't satisfy the stoichiometry and volume change.

Key Concepts:
- Stoichiometry
- Gas reactions
- Volume changes

Memory Aids:
- Remember to balance the chemical equation.

Quick Tips:
- Use stoichiometry to relate the volumes of reactants and products.

2 Medium

2 of 272

Application +4 | -1

How many millilitre of

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

are needed to dissolve

0.5 \mathrm{~g}

of

\mathrm{Cu}(\mathrm{II})

carbonate?

View answer and explanation

Correct option: C

8.1

Option C (8.1) is correct because it is calculated using the stoichiometry of the reaction between H2SO4 and CuCO3.

Solution approach

Correct Answer + Core Reason: 8.1. The volume of H2SO4 needed is calculated using stoichiometry.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $CuCO_3 + H_2SO_4 \to CuSO_4 + H_2O + CO_2$

Step 2: Calculate the number of moles of $CuCO_3$ : Molar mass of $CuCO_3 = 63.5 + 12 + 3(16) = 123.5 g/mol$ . Moles of $CuCO_3 = \frac{{0.5 g}}{{123.5 g/mol}} = 0.00405 mol$

Step 3: From the balanced equation, 1 mole of $CuCO_3$ reacts with 1 mole of $H_2SO_4$ . Therefore, moles of $H_2SO_4$ needed = 0.00405 mol.

Step 4: Use the molarity of $H_2SO_4$ to find the volume: $Molarity = \frac{{moles}}{{Volume (L)}}$ . $0.5 M = \frac{{0.00405 mol}}{{Volume (L)}}$ . Volume (L) = $\frac{{0.00405 mol}}{{0.5 mol/L}} = 0.0081 L$

Step 5: Convert the volume to milliliters: Volume (mL) = 0.0081 L * 1000 mL/L = 8.1 mL

Why other options are wrong:
- Option A (6.01): This is incorrect because of a miscalculation in the stoichiometry or molarity calculation.
- Option B (4.5): This is incorrect because of a miscalculation in the stoichiometry or molarity calculation.
- Option D (11.1): This is incorrect because of a miscalculation in the stoichiometry or molarity calculation.

Key Concepts:
- Stoichiometry
- Molarity
- Balanced chemical equations

Memory Aids:
- Molarity = moles/volume

Quick Tips:
- Always write a balanced chemical equation first.

3 Medium

3 of 272

Understanding +4 | -1

Statement I: Chemical equivalent of

\mathrm{Mg}

is half of its atomic weight

Statement II: Charge present on

\mathrm{Mg}

is +2

View answer and explanation

Correct option: A

If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Option A (If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because the chemical equivalent of an element is its atomic weight divided by its valency, and the valency of Mg is 2.

Solution approach

Correct Answer + Core Reason: If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I. The equivalent weight is the atomic weight divided by the valency.

Step-by-Step Solution:
Step 1: Define chemical equivalent: The chemical equivalent of an element is its atomic weight divided by its valency.
Step 2: State Statement I: Chemical equivalent of $\mathrm{Mg}$ is half of its atomic weight. This is true because the valency of Mg is 2.
Step 3: State Statement II: Charge present on $\mathrm{Mg}$ is +2. This is true because magnesium is in Group 2 of the periodic table and forms a +2 ion.
Step 4: Determine if Statement II is the correct explanation of Statement I. Since the charge on Mg is +2, its valency is 2. Therefore, its equivalent weight is its atomic weight divided by 2, which is half of its atomic weight.

Why other options are wrong:
- Option B: If both Statement I and Statement II are correct and Statement II is not a correct explanation of Statement I is incorrect because Statement II is the correct explanation.
- Option C: Statement I is correct but statement II is false is incorrect because both statements are correct.
- Option D: Statement I is false but Statement II is correct is incorrect because both statements are correct.

Key Concepts:
- Chemical equivalent
- Valency
- Atomic weight

Memory Aids:
- Remember that equivalent weight = atomic weight / valency.

Quick Tips:
- Relate the charge of an ion to its valency.

4 Medium

4 of 272

Application +4 | -1

If the mass of neutron is doubled and that of proton is halved, the molecular mass of

\mathrm{H}_2 \mathrm{O}

containing only

\mathrm{H}^1

and

\mathrm{O}^{16}

atoms will

View answer and explanation

Correct option: C

Increase by about

16.67 \%

Option C (Increase by about 16.67 %) is correct because the new molecular mass of H2O increases from 18 to 21, which is approximately a 16.67% increase.

Solution approach

Correct Answer + Core Reason: Increase by about 16.67 %. The molecular mass increases due to the change in neutron and proton masses.

Step-by-Step Solution:

Step 1: Original mass of neutron = 1, New mass of neutron = 2

Step 2: Original mass of proton = 1, New mass of proton = 0.5

Step 3: Original molecular mass of $H_2O = 2(1) + 16 = 18$

Step 4: New molecular mass of $H_2O = 2(0.5) + 16 = 1 + 16 + 4 = 21$

Step 5: Calculate the percentage increase: $\frac{{New Mass - Original Mass}}{{Original Mass}} \times 100 = \frac{{21 - 18}}{{18}} \times 100 = \frac{3}{{18}} \times 100 = 16.67\%$

Why other options are wrong:
- Option A (Increase by about 25 %): This is incorrect because the percentage increase is not 25%.
- Option B (Decrease by about 25 %): This is incorrect because the molecular mass increases, not decreases.
- Option D (Decrease by about 16.67 %): This is incorrect because the molecular mass increases, not decreases.

Key Concepts:
- Molecular mass
- Atomic mass
- Percentage change

Memory Aids:
- Molecular mass is the sum of atomic masses.

Quick Tips:
- Carefully consider the changes in atomic masses when calculating the new molecular mass.

5 Medium

5 of 272

Analyzing +4 | -1

A: Number of atoms in 2 mole of

{NH}_3

is equal to number of atoms in 4 mole of

{CH}_4

.

R: Both are chemically similar species.

View answer and explanation

Correct option: D

Both the A and R are incorrect.

Option D is correct because the number of atoms in 2 moles of NH3 is not equal to the number of atoms in 4 moles of CH4, and they are chemically different.

Solution approach

Correct Answer + Core Reason: Both the A and R are incorrect. The number of atoms and chemical similarity are both incorrect.

Step-by-Step Solution:

Step 1: Calculate the number of atoms in 2 moles of NH3.\nNumber of atoms in 1 molecule of NH3 = 1 N atom + 3 H atoms = 4 atoms\nNumber of atoms in 1 mole of NH3 = 4 * Avogadro's number ( $\mathrm{N_0}$ )\nNumber of atoms in 2 moles of NH3 = 2 * 4 * $\mathrm{N_0}$ = 8 $\mathrm{N_0}$

Step 2: Calculate the number of atoms in 4 moles of CH4.\nNumber of atoms in 1 molecule of CH4 = 1 C atom + 4 H atoms = 5 atoms\nNumber of atoms in 1 mole of CH4 = 5 * Avogadro's number ( $\mathrm{N_0}$ )\nNumber of atoms in 4 moles of CH4 = 4 * 5 * $\mathrm{N_0}$ = 20 $\mathrm{N_0}$

Step 3: Compare the number of atoms in 2 moles of NH3 and 4 moles of CH4.
8 $\mathrm{N_0}$ is not equal to 20 $\mathrm{N_0}$

Step 4: Determine if NH3 and CH4 are chemically similar species.\nNH3 (ammonia) and CH4 (methane) are chemically different species. Ammonia is a polar molecule with a lone pair of electrons on the nitrogen atom, while methane is a nonpolar molecule.

Step 5: Conclude that both the assertion and the reason are incorrect.

Why other options are wrong:
- Option A: Both A and R are correct and R is the correct explanation of A. This is incorrect because both A and R are incorrect.
- Option B: Both A and R are correct, but R is not the correct explanation of A. This is incorrect because both A and R are incorrect.
- Option C: A is correct but R is incorrect. This is incorrect because both A and R are incorrect.

Key Concepts:
- Mole concept
- Avogadro's number
- Chemical similarity

Memory Aids:
- Remember that 1 mole of any substance contains Avogadro's number of particles.

Quick Tips:
- Always calculate the number of atoms or molecules based on the number of moles and Avogadro's number.
- Understand the chemical properties of different species to determine if they are chemically similar.

6 Medium

6 of 272

Application +4 | -1

In the reaction,

2\mathrm{CuS}{\mathrm{O}_4} + 4\mathrm{KI} \to \mathrm{C}{\mathrm{u}_2}{\mathrm{I}_2} + 2{\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} + {\mathrm{I}_2}

The ratio of equivalent weight of

\mathrm{CuS}{\mathrm{O}_4}

to its molecular weight is :

View answer and explanation

Correct option: D

1

Option D (1) is correct because each Cu2+ ion is reduced to Cu+ in Cu2I2, involving a change of 1 electron per copper atom, thus the equivalent weight is equal to the molecular weight.

Solution approach

Correct Answer + Core Reason: 1. The ratio of equivalent weight to molecular weight is 1 because each copper ion undergoes a change of 1 in oxidation state.

Step-by-Step Solution:

Step 1: Identify the change in oxidation state: $2\mathrm{C}{\mathrm{u}^{2 + }} + 2e \to \mathrm{Cu}_2^{1 + }$

Step 2: Determine the n-factor: The n-factor is 1 because each copper ion gains 1 electron.

Step 3: Calculate the equivalent weight: $E = \frac{M}{n} = \frac{M}{1} = M$

Step 4: Calculate the ratio: $\frac{E}{M} = \frac{M}{M} = 1$

Why other options are wrong:
- Option A: 1/8 is incorrect because it does not represent the correct change in oxidation state of copper.
- Option B: 1/4 is incorrect because it does not represent the correct change in oxidation state of copper.
- Option C: 1/82 is incorrect because it does not represent the correct change in oxidation state of copper.

Key Concepts:
- Equivalent weight
- Oxidation-reduction reactions

Memory Aids:
- Equivalent weight = Molecular weight / n-factor

Quick Tips:
- Determine the change in oxidation state to find the n-factor.
- Remember the formula for equivalent weight.

7 Medium

7 of 272

Analyzing +4 | -1

A drug, marijuana, owes its activity to tetrahydrocannabinol, which contains 70 per cent as many carbon atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. The number of moles in a gram of tetrahydrocannabinol is 0.00318. Determine its molecular formula.

View answer and explanation

Correct option: B

\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_2

Option B (C21H30O2) is correct because the molecular formula is determined by using the given ratios of C, H, and O and the number of moles in a gram of the compound.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_2$ . The molecular formula is determined from the given ratios and moles per gram.

Step-by-Step Solution:
Step 1: Define the number of atoms: Let the number of oxygen atoms be $x$ . Then, the number of hydrogen atoms is $15x$ , and the number of carbon atoms is $0.70 \times 15x = 10.5x$ .
Step 2: Write the general molecular formula: $\mathrm{C}_{10.5x} \mathrm{H}_{15x} \mathrm{O}_x$ .
Step 3: Calculate the molar mass: The molar mass is $12(10.5x) + 1(15x) + 16(x) = 126x + 15x + 16x = 157x$ g/mol.
Step 4: Use the given information about moles per gram: $\frac{1 \text{ g}}{157x \text{ g/mol}} = 0.00318 \text{ mol}$ .
Step 5: Solve for $x$ : $1 = 0.00318 \times 157x$ , so $x = \frac{1}{0.00318 \times 157} \approx 2$ .
Step 6: Determine the molecular formula: $\mathrm{C}_{10.5(2)} \mathrm{H}_{15(2)} \mathrm{O}_2 = \mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_2$ .

Why other options are wrong:
- Option A: $\mathrm{CH}_3 \mathrm{O}_2$ : This does not match the given ratios.
- Option C: $\mathrm{C}_{15} \mathrm{H}_{30} \mathrm{O}_2$ : This does not match the given ratios.
- Option D: $\mathrm{C}_{70} \mathrm{H}_{15} \mathrm{O}$ : This does not match the given ratios.

Key Concepts:
- Molecular formula determination
- Molar mass
- Mole concept

Memory Aids:
- Use the given ratios to relate the number of atoms.

Quick Tips:
- Always check if the calculated molecular formula makes sense based on the given information.

8 Medium

8 of 272

Understanding +4 | -1

Assertion: Equivalent weight of

\mathrm{KMnO}_4

in acidic medium is

\frac{\mathrm{M}}{5}

.
Reason: In acidic medium 1 mol of

\mathrm{MnO}_4^{-}

gains 5 electron.

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion

Option A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) is correct because in acidic medium, permanganate ion (

MnO_4^−

) gains 5 electrons, and this directly explains why the equivalent weight is M/5.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. The equivalent weight is M/5 because 1 mol of $MnO_4^-$ gains 5 electrons in acidic medium.

Step-by-Step Solution:

Step 1: Recall the definition of equivalent weight.

Step 2: Understand the redox reaction of $KMnO_4$ in acidic medium.
$\mathop {{\text{Mn}}}\limits^{ + 7} {\text{O}}_4^ - \stackrel{{+5 \mathrm{e}^{-}}}{\longrightarrow} \mathrm{Mn}^{2+}$

Step 3: Determine the number of electrons gained by $MnO_4^-$ in acidic medium. It gains 5 electrons.

Step 4: Calculate the equivalent weight of $KMnO_4$ in acidic medium.\nEquivalent weight = $\frac{{\text{Molecular weight}}}{{\text{n-factor}}}$ = $\frac{{\text{M}}}{5}$

Why other options are wrong:
- Option B: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion - Reason is the correct explanation.
- Option C: If Assertion is correct but Reason is incorrect - Reason is correct.
- Option D: If both Assertion and Reason are incorrect - Both are correct.

Key Concepts:
- Equivalent weight
- Redox reactions
- n-factor

Memory Aids:
- Remember the n-factor of $KMnO_4$ in different media.

Quick Tips:
- Determine the n-factor based on the number of electrons gained or lost.

9 Medium

9 of 272

Understanding +4 | -1

Which term is to be correctly used for expressing concentration of electrolytes in solution?

View answer and explanation

Correct option: C

Formality

Option C (Formality) is correct because formality is specifically used to express the concentration of electrolytes in solution, accounting for their dissociation.

Solution approach

Correct Answer + Core Reason: Formality. Formality is used for expressing the concentration of electrolytes in solution.

Step-by-Step Solution:

Step 1: Define the concentration terms:
- Molarity: Moles of solute per liter of solution.
- Normality: Gram equivalent weight of solute per liter of solution.
- Formality: Gram formula weight of solute per liter of solution.

Step 2: Understand the application of each term:
- Molarity is used for molecular compounds.
- Normality is used in acid-base titrations and redox reactions.
- Formality is used for ionic compounds (electrolytes) because they exist as ions in solution, and their exact molecular weight is difficult to determine experimentally.

Step 3: Conclusion: Formality is the most appropriate term for expressing the concentration of electrolytes in solution.

Why other options are wrong:
- Option A: Molarity is generally used for molecular compounds, not electrolytes.
- Option B: Normality is used in specific types of reactions, not as a general concentration term for electrolytes.
- Option D: Formality is the correct answer.

Key Concepts:
- Concentration terms (Molarity, Normality, Formality)
- Electrolytes
- Ionic compounds

Memory Aids:
- Formality = Formula weight = For Ionic compounds.

Quick Tips:
- Remember that formality is specifically used for ionic compounds (electrolytes).

10 Medium

10 of 272

Analyzing +4 | -1

A: One mole of sucrose reacts completely with oxygen produces 268.8 litre of carbon dioxide at STP.

R: Amount of oxygen required for reaction is 268.8 litre.

View answer and explanation

Correct option: B

Both A and R are correct, but R is not the correct explanation of A.

Both the assertion and the reason are correct, but the reason is not the correct explanation of the assertion. The amount of CO2 produced and O2 required are both 268.8 L, but the production of CO2 is not *caused* by the amount of oxygen required.

Solution approach

Correct Answer + Core Reason: Both A and R are correct, but R is not the correct explanation of A. Both the volume of CO2 produced and the volume of O2 required are 268.8 L, but they are independent.

Step-by-Step Solution:
Step 1: Write the balanced chemical equation: C12H22O11(s) + 12O2(g) -> 12CO2(g) + 11H2O(l)
Step 2: Analyze the stoichiometry: 1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of carbon dioxide.
Step 3: Calculate the volume of CO2 produced: 12 moles of CO2 at STP occupy 12 * 22.4 L = 268.8 L.
Step 4: Calculate the volume of O2 required: 12 moles of O2 at STP occupy 12 * 22.4 L = 268.8 L.
Step 5: Evaluate the statements: Both the assertion and the reason are correct. However, the reason is not the correct explanation for the assertion.

Why other options are wrong:
- Option A: Both A and R are correct and R is the correct explanation of A.: This is incorrect because the reason is not the correct explanation.
- Option C: A is correct but R is incorrect.: This is incorrect because both A and R are correct.
- Option D: Both the A and R are incorrect.: This is incorrect because both A and R are correct.

Key Concepts:
- Stoichiometry
- Balanced chemical equations
- Molar volume of gases at STP

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

Quick Tips:
- Always write a balanced chemical equation before performing stoichiometric calculations.

11 Medium

11 of 272

Application +4 | -1

What is the equivalent weight of

{ClO}_3^{-}

in the following reaction?

{ClO}_3^{-}+{Fe}^{2+}+{H}^{+} \longrightarrow {Cl}^{-}+{Fe}^{3+}+{H}_2 {O}

View answer and explanation

Correct option: A

13.9167

Option A (13.9167) is correct because it is the equivalent weight of ClO₃⁻ calculated based on the change in oxidation state of chlorine.

Solution approach

Correct Answer + Core Reason: 13.9167. The equivalent weight of ClO3- is calculated based on the change in oxidation state of chlorine.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Cl in $ClO_3^-$ : The oxidation state of O is -2, so the oxidation state of Cl is +5.

Step 2: Determine the oxidation state of Cl in $Cl^-$ : The oxidation state of Cl is -1.

Step 3: Calculate the change in oxidation state: Change = +5 - (-1) = 6

Step 4: Calculate the equivalent weight: Equivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{Change in oxidation state}}}} = \frac{{(35.5 + 3 \times 16)}}{{6}} = \frac{{83.5}}{{6}} = 13.9167$

Why other options are wrong:
- Option B (12.9): This is incorrect. It is likely a calculation error or misunderstanding of the concept of equivalent weight.
- Option C (11.9): This is incorrect. It might be due to an incorrect calculation of the change in oxidation state.
- Option D (10.9): This is incorrect. It is likely a calculation error in the formula.

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

Memory Aids:
- Equivalent weight

12 Medium

12 of 272

Understanding +4 | -1

The equivalent weight of an acid is obtained by dividing its mol. wt. by its:

View answer and explanation

Correct option: B

Basicity

Option B is correct because the equivalent weight of an acid is determined by dividing its molecular weight by its basicity (number of replaceable hydrogen ions).

Solution approach

Correct Answer + Core Reason: Basicity. The equivalent weight of an acid is calculated by dividing its molecular weight by its basicity.

Step-by-Step Solution:
Step 1: Recall the definition of equivalent weight.
Step 2: Understand that for acids, equivalent weight = (Molecular weight) / (Basicity).

Why other options are wrong:
- Option A: Acidity is a property of a solution, not a characteristic of the acid itself used to calculate equivalent weight.
- Option C: pH is a measure of the acidity of a solution, not used to calculate equivalent weight.
- Option D: Basicity is the correct factor to divide the molecular weight by.

Key Concepts:
- Equivalent Weight
- Basicity of an Acid
- Molecular Weight

Memory Aids:
- Equivalent weight = Molecular weight / Basicity (for acids)

Quick Tips:
- Remember that basicity refers to the number of replaceable hydrogen ions in an acid.

13 Medium

13 of 272

Application +4 | -1

The no. of moles of Hydrogen molecules required to produce 20 moles of

N{H_3}

through Haber’s process is

View answer and explanation

Correct option: B

30

Option B (30) is correct because the balanced equation for Haber's process shows that 3 moles of H2 are required for every 2 moles of NH3.

Solution approach

Correct Answer + Core Reason: 30. The number of moles of hydrogen is determined from the stoichiometry of Haber's process.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for Haber's process.
${N_2} + 3{H_2} \to 2N{H_3}$

Step 2: Determine the mole ratio between ${H_2}$ and $N{H_3}$ .\nFrom the balanced equation, 3 moles of ${H_2}$ produce 2 moles of $N{H_3}$ .

Step 3: Calculate the number of moles of ${H_2}$ required to produce 20 moles of $N{H_3}$ .\nMoles of ${H_2}$ = $\frac{3}{2} \times Moles\,of\,N{H_3}$ \nMoles of ${H_2}$ = $\frac{3}{2} \times 20 = 30$

Why other options are wrong:
- Option A (20): This is incorrect because it doesn't account for the stoichiometric ratio between hydrogen and ammonia.
- Option C (40): This is incorrect and suggests an error in the stoichiometric calculation.
- Option D (10): This is incorrect and represents a miscalculation in the mole ratio.

Key Concepts:
- Stoichiometry
- Haber's process
- Mole ratio

Memory Aids:
- Balanced chemical equation is essential for stoichiometric calculations.

Quick Tips:
- Always start with a balanced chemical equation.
- Use the mole ratio to convert between reactants and products.

14 Medium

14 of 272

Application +4 | -1

The number of electrons in

2 \mathrm{~g}

ion of nitrate ion

\left(\mathrm{NO}_3^{-}\right)

is

View answer and explanation

Correct option: B

64 N_{\mathrm{A}}

Option B (

64 N_{\mathrm{A}}

) is correct because the total number of electrons is calculated by multiplying the number of nitrate ions by the number of electrons per nitrate ion.

Solution approach

Correct Answer + Core Reason: $64 N_{\mathrm{A}}$ . The total number of electrons is calculated by multiplying the number of nitrate ions by the number of electrons per nitrate ion.

Step-by-Step Solution:

Step 1: Calculate the number of electrons in one nitrate ion ( $\mathrm{NO}_3^{-}$ ).\nNitrogen (N) has 7 electrons.\nOxygen (O) has 8 electrons.\nThe negative charge adds 1 electron.\nTotal electrons in $\mathrm{NO}_3^{-}$ = 7 + 3(8) + 1 = 7 + 24 + 1 = 32 electrons.

Step 2: Calculate the number of nitrate ions in 2 g-ions of nitrate.
1 g-ion of $\mathrm{NO}_3^{-}$ contains $N_{\mathrm{A}}$ ions, where $N_{\mathrm{A}}$ is Avogadro's number.
2 g-ions of $\mathrm{NO}_3^{-}$ contain $2N_{\mathrm{A}}$ ions.

Step 3: Calculate the total number of electrons.\nTotal number of electrons = (Number of electrons per ion) * (Number of ions)\nTotal number of electrons = 32 * $2N_{\mathrm{A}}$ = $64N_{\mathrm{A}}$

Why other options are wrong:
- Option A (64): This is incorrect because it only gives the number of electrons per 2 g-ions without multiplying by Avogadro's number.
- Option C (32): This is incorrect because it only gives the number of electrons in one nitrate ion.
- **Option D ( $32 N_{\mathrm{A}}$ ): This is incorrect because it only gives the number of electrons in 1 g-ion of nitrate ions.

Key Concepts:
- Number of Electrons in Ions
- Avogadro's Number
- Gram-ion

Memory Aids:
- Remember that 1 g-ion contains Avogadro's number of ions.

Quick Tips:**
- Always consider the charge of the ion when calculating the number of electrons.
- Remember to multiply by Avogadro's number when dealing with g-ions or moles.

15 Medium

15 of 272

Understanding +4 | -1

Irrespective of the source, pure sample, of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of

View answer and explanation

Correct option: C

constant composition

The correct answer is C (constant composition) because the fixed mass ratio of oxygen and hydrogen in water, regardless of its source, exemplifies the law of constant composition.

Solution approach

Correct Answer + Core Reason: constant composition. The fixed ratio of elements in a compound is the basis of the law of constant composition.

Step-by-Step Solution:

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

Step 2: State the law of multiple proportions: 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 3: State the law of constant composition: A given compound always contains exactly the same proportion of elements by mass.

Step 4: State the law of constant volume: At constant temperature and pressure, the volumes of reacting gases and gaseous products bear a simple whole number ratio.

Step 5: Apply the laws to the given information: The fixed mass ratio of oxygen and hydrogen in water demonstrates the law of constant composition.

Why other options are wrong:
- Option A (conservation of mass): This law deals with the conservation of mass during a chemical reaction, not the composition of a compound.
- Option B (multiple proportions): This law applies when two elements form multiple compounds, not a single compound like water.
- Option D (constant volume): This law applies to gases, not the composition of a compound.

Key Concepts:
- Laws of chemical combination
- Law of constant composition

Memory Aids:
- Constant composition = fixed ratio of elements.

Quick Tips:
- Understand the definitions of the different laws of chemical combination.

16 Medium

16 of 272

Application +4 | -1

One molecule of haemoglobin will combine with four molecules of oxygen. If

1.0 \mathrm{~g}

of haemoglobin combines with

1.642 \mathrm{~ml}

of oxygen at body temperature

\left(27^{\circ} \mathrm{C}\right)

and a pressure of 760 torr, then what is the molar mass of haemoglobin?

View answer and explanation

Correct option: D

60,000

Option D (60,000) is correct because the molar mass of hemoglobin is calculated using the ideal gas law and the stoichiometry of oxygen binding to hemoglobin.

Solution approach

Correct Answer + Core Reason: 60,000. The molar mass is calculated using the ideal gas law and stoichiometry.

Step-by-Step Solution:

Step 1: Write the ideal gas law: $PV = nRT$ , where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

Step 2: Convert the given values to appropriate units: V = 1.642 mL = 0.001642 L, T = 27°C = 300 K, P = 760 torr = 1 atm, R = 0.0821 L atm K⁻¹ mol⁻¹.

Step 3: Calculate the number of moles of oxygen: $n = \frac{{PV}}{{RT}} = \frac{{1 \times 0.001642}}{{0.0821 \times 300}} = 6.67 \times 10^{-5} \text{{ mol}}$

Step 4: Use the stoichiometry of the reaction. One molecule of hemoglobin combines with four molecules of oxygen. Therefore, the number of moles of hemoglobin is $\frac{{1}}{{4}}$ the number of moles of oxygen.

Step 5: Calculate the number of moles of hemoglobin: $n_{Hb} = \frac{{6.67 \times 10^{-5}}}{{4}} = 1.6675 \times 10^{-5} \text{{ mol}}$

Step 6: Calculate the molar mass of hemoglobin: Molar mass = $\frac{{\text{{mass}}}}{{\text{{moles}}}} = \frac{{1.0 \text{{ g}}}}{{1.6675 \times 10^{-5} \text{{ mol}}}} = 60000 \text{{ g/mol}}$

Why other options are wrong:
- Option A (6,00,000): This is incorrect due to a miscalculation in the molar mass, possibly by not dividing by 4 for the stoichiometry.
- Option B (1,50,000): This is incorrect due to a miscalculation in the molar mass, possibly by not dividing by 4 for the stoichiometry or using incorrect values.
- Option C (15,000): This is incorrect due to a miscalculation in the molar mass, possibly by not dividing by 4 for the stoichiometry or using incorrect values.

Key Concepts:
- Ideal gas law
- Stoichiometry
- Molar mass

Memory Aids:
- PV = nRT

Quick Tips:
- Remember to convert all units to be consistent.
- Pay attention to the stoichiometry of the reaction.

17 Medium

17 of 272

Application +4 | -1

6.02 \times 10^{20}

molecules of urea are present in

200 \mathrm{~mL}

of its solution. The concentration of urea solution is

View answer and explanation

Correct option: C

0.005 \mathrm{~M}

Option C (

0.005 \mathrm{~M}

) is correct because the concentration is calculated by dividing the number of moles of urea by the volume of the solution in liters.

Solution approach

Correct Answer + Core Reason: $0.005 \mathrm{~M}$ . The concentration of the urea solution is calculated by dividing the number of moles by the volume in liters.

Step-by-Step Solution:

Step 1: Calculate the number of moles of urea.
$\text {Moles of urea}=\frac{6.02 \times 10^{20}}{6.02 \times 10^{23}}=10^{-3} \text { moles }$

Step 2: Calculate the molarity of the urea solution.
$\text {Molarity of urea}=\frac{10^{-3} \times 1000}{200}=0.005 \mathrm{~M}$

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

Key Concepts:
- Molarity calculation
- Mole concept

Memory Aids:
- Remember that Molarity = Moles / Volume (in Liters)

Quick Tips:
- Ensure the volume is in liters before calculating molarity.

18 Medium

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Application +4 | -1

The mass of

1.5 \times 10^{20}

molecules of a substance is

20 \mathrm{~mg}

. The molar mass of the substance is

View answer and explanation

Correct option: C

80 \mathrm{~g}

Option C (

80 \mathrm{~g}

) is correct because it is the molar mass calculated from the given number of molecules and mass of the substance.

Solution approach

Correct Answer + Core Reason: $80 \mathrm{~g}$ . The molar mass is calculated using the given mass and number of molecules, scaled to Avogadro's number.

Step-by-Step Solution:

Step 1: Given: $1.5 \times 10^{20}$ molecules have a mass of $20 \mathrm{~mg} = 20 \times 10^{-3} \mathrm{~g}$ .

Step 2: Find the mass of $6.022 \times 10^{23}$ molecules (1 mole).

Step 3: Set up a proportion: $\frac{{1.5 \times 10^{20} \text{{ molecules}}}}{{20 \times 10^{-3} \text{{ g}}}} = \frac{{6.022 \times 10^{23} \text{{ molecules}}}}{{x \text{{ g}}}}$

Step 4: Solve for x: $x = \frac{{20 \times 10^{-3} \times 6.022 \times 10^{23}}}{{1.5 \times 10^{20}}} = \frac{{120.44 \times 10^{20}}}{{1.5 \times 10^{20}}} = 80.29 \approx 80 \mathrm{~g}$ .

Why other options are wrong:
- **Option A ( $20 \mathrm{~g}$ ): This is incorrect as it simply states the mass of the given sample without scaling to a mole.
- Option B ( $40 \mathrm{~g}$ ): This is incorrect and likely arises from a calculation error in the scaling process.
- Option D ( $80 \mathrm{~a.m.u}$ ): This is incorrect because molar mass should be in grams per mole (g/mol), not atomic mass units (amu).

Key Concepts:**
- Mole concept
- Avogadro's number ( $6.022 \times 10^{23}$ )
- Molar mass

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

Quick Tips:
- Always convert mass to grams before calculations.
- Ensure units are consistent throughout the calculation.

19 Medium

19 of 272

Analyzing +4 | -1

In the reaction of oxalate with permanganate in acidic medium. The number of electrons involved in producing one molecule of

C{O_2}

is

View answer and explanation

Correct option: C

1

Option C is correct because the oxidation of one oxalate ion to two CO2 molecules involves a transfer of 2 electrons, meaning 1 electron per CO2 molecule.

Solution approach

Correct Answer + Core Reason: 1. One electron is involved in producing one molecule of CO2.

Step-by-Step Solution:

Step 1: Write the balanced half-reaction for the oxidation of oxalate to carbon dioxide.
- $\mathrm{C_2O_4^{2-} \longrightarrow 2CO_2 + 2e^-}$

Step 2: Analyze the stoichiometry of the half-reaction.
- One oxalate ion ( $\mathrm{C_2O_4^{2-}}$ ) produces two carbon dioxide molecules ( $\mathrm{2CO_2}$ ) and releases two electrons ( $\mathrm{2e^-}$ ).

Step 3: Calculate the number of electrons involved in producing one molecule of carbon dioxide.
- Since two electrons are released for two molecules of carbon dioxide, one electron is involved in producing one molecule of carbon dioxide.

Why other options are wrong:
- Option A: 2 electrons are involved in producing 2 molecules of CO2, not 1 molecule.
- Option B: 5 is the number of electrons gained by MnO4- in the redox reaction, not related to CO2 production directly.
- Option D: 10 is the total number of electrons transferred in the balanced redox reaction with 5 oxalate ions, not related to CO2 production directly.

Key Concepts:
- Redox Reactions
- Half-Reactions
- Stoichiometry

Memory Aids:
- Remember to balance the half-reactions to determine the number of electrons transferred.

Quick Tips:
- Write the balanced half-reaction to determine the number of electrons involved.

20 Medium

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Application +4 | -1

Manganese forms non-stoichiometric oxides having the general formula

\mathrm{MnO}_x

. The value of

x

for the compound that analyzed

64 \%

by mass

\mathrm{Mn}

:

View answer and explanation

Correct option: D

1.93

Option D (1.93) is correct because the value of x is calculated using the given mass percentage of Mn in the non-stoichiometric oxide.

Solution approach

Correct Answer + Core Reason: 1.93. The value of x is calculated from the mass percentage of Mn.

Step-by-Step Solution:

Step 1: Set up the equation for the mass percentage of Mn in MnOx.

Step 2: The formula for the mass percentage of Mn is: $% \text{{Mn}} = \frac{{\text{{Molar mass of Mn}}}}{{\text{{Molar mass of Mn}} + x \times \text{{Molar mass of O}}}} \times 100$

Step 3: Substitute the molar masses: $% \text{{Mn}} = \frac{{55}}{{55 + 16x}} \times 100$

Step 4: Set the mass percentage equal to 64: $\frac{{55}}{{55 + 16x}} \times 100 = 64$

Step 5: Solve for x: $5500 = 64(55 + 16x) \Rightarrow 5500 = 3520 + 1024x \Rightarrow 1980 = 1024x \Rightarrow x = \frac{{1980}}{{1024}} = 1.93$

Step 6: Final Answer: $x = 1.93$

Why other options are wrong:
- Option A (1.16): This value is incorrect because it results from an incorrect calculation or setup of the mass percentage equation.
- Option B (1.83): This value is incorrect due to a calculation error in solving for x.
- Option C (2): This value would represent stoichiometric MnO2, but the problem specifies a non-stoichiometric oxide, so this is incorrect.

Key Concepts:
- Mass percentage calculation
- Non-stoichiometric compounds

Memory Aids:
- Remember the formula: % element = (mass of element / mass of compound) * 100

Quick Tips:
- Always double-check your algebra when solving for x in these types of problems.

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Analyzing +4 | -1

Assertion (A): On changing volume of the solution by

20 \%

, molarity of solution also changes by

20 \%

.

Reason (R): Molar concentration or molarity of solution is not affected on dilution.

View answer and explanation

Correct option: D

Both A and R are false

Option D (Both A and R are false) is correct because molarity changes inversely with volume, and the percentage change in molarity is not equal to the percentage change in volume. Also, molarity *is* affected by dilution.

Solution approach

Correct Answer + Core Reason: Both A and R are false. Molarity changes inversely with volume, and molarity is affected by dilution.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): On changing the volume of the solution by 20%, the molarity of the solution also changes by 20%.

Step 2: Let the initial volume be V and the initial molarity be M. The number of moles of solute remains constant.

Step 3: If the volume changes by 20%, the new volume is 1.2V.

Step 4: The new molarity (M') can be calculated using the formula: M1V1 = M2V2. MV = M'(1.2V), so M' = M/1.2 = 0.833M.

Step 5: The percentage change in molarity is (M - 0.833M)/M * 100 = 16.67%, which is not 20%. Therefore, Assertion (A) is false.

Step 6: Analyze Reason (R): Molar concentration or molarity of the solution is not affected by dilution.

Step 7: Dilution involves adding more solvent to the solution, which changes the volume and therefore the molarity. Therefore, Reason (R) is false.

Why other options are wrong:
- Option A: Both A and R are true and R is the correct explanation of A: Both A and R are false.
- Option B: Both A and R are true but R is not the correct explanation of A: Both A and R are false.
- Option C: A is true but R is false: A is also false.

Key Concepts:
- Molarity
- Dilution
- Moles

Memory Aids:
- Molarity = Moles of solute / Volume of solution

Quick Tips:
- Remember that molarity is inversely proportional to volume during dilution.

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Application +4 | -1

Some bottles of colourless liquids were being labelled when the technicians accidentally mixed them up and lost track of their contents. A

15.0 \mathrm{~ml}

sample withdrawn from one bottle weighed

22.3 \mathrm{~g}.

The technicians knew that the liquid was either acetone, benzene, chloroform or carbon tetrachloride (which have densities of

0.792 \mathrm{~g} /\mathrm{cm}^3, 0.899 \mathrm{~g} / \mathrm{cm}^3, 1.489 \mathrm{~g} / \mathrm{cm}^3

, and

1.595 \mathrm{~g} / \mathrm{cm}^3

, respectively). What was the identity of the liquid?

View answer and explanation

Correct option: C

Chloroform

Option C (Chloroform) is correct because the calculated density of the liquid matches the density of chloroform.

Solution approach

Correct Answer + Core Reason: Chloroform. The density calculated from the given mass and volume matches that of chloroform.

Step-by-Step Solution:

Step 1: Calculate the density of the liquid using the given mass and volume: $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$

Step 2: Substitute the given values: $\text{Density} = \frac{22.3 \text{ g}}{15.0 \text{ mL}} = 1.48666... \text{ g/mL} \approx 1.487 \text{ g/mL}$

Step 3: Compare the calculated density with the given densities of the substances.

Step 4: The calculated density (1.487 g/mL) is closest to the density of chloroform (1.489 g/mL).

Why other options are wrong:
- Option A (Carbon tetrachloride): The density of carbon tetrachloride (1.595 g/mL) is significantly different from the calculated density.
- Option B (Acetone): The density of acetone (0.792 g/mL) is significantly different from the calculated density.
- Option D (Benzene): The density of benzene (0.899 g/mL) is significantly different from the calculated density.

Key Concepts:
- Density calculation
- Identification of substances based on density

Memory Aids:
- Density = Mass / Volume

Quick Tips:
- Calculate the density and compare it with the known densities of the possible substances.

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Application +4 | -1

A sample of ammonium phosphate,

{\left( {N{H_4}} \right)_3}P{O_4}

, contain 6 mole of hydrogen atoms. The number of moles of oxygen atoms in the sample is

View answer and explanation

Correct option: B

2

Option B (2) is correct because in one mole of ammonium phosphate, 12 moles of hydrogen atoms correspond to 4 moles of oxygen atoms, thus 6 moles of hydrogen correspond to 2 moles of oxygen.

Solution approach

Correct Answer + Core Reason: 2. The number of moles of oxygen atoms is calculated based on the ratio of hydrogen to oxygen in ammonium phosphate.

Step-by-Step Solution:

Step 1: Analyze the formula of ammonium phosphate.
${(N{H_4})_3}P{O_4} \to$ 12 moles of H – atoms & 4 moles of O – atoms

Step 2: Determine the ratio of hydrogen to oxygen.\nFor every 12 moles of H, there are 4 moles of O.

Step 3: Calculate the moles of oxygen for 6 moles of hydrogen.\nIf 12 moles of H correspond to 4 moles of O, then 6 moles of H correspond to (6/12) * 4 = 2 moles of O.

Why other options are wrong:
- Option A: Incorrect calculation of the moles of oxygen.
- Option C: Incorrect calculation of the moles of oxygen.
- Option D: Incorrect calculation of the moles of oxygen.

Key Concepts:
- Mole concept
- Chemical formula interpretation

Memory Aids:
- Remember to analyze the chemical formula carefully to determine the ratios of atoms.

Quick Tips:
- Always check the stoichiometry of the compound to find the correct ratio.

24 Medium

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Application +4 | -1

In a chemical reaction,

A + B \to AB

if B is acting as a limiting reagent then choose the correct option

View answer and explanation

Correct option: C

A, 50 atoms; B, 30 atoms

Option C (A, 50 atoms; B, 30 atoms) is correct because B is the limiting reagent, meaning there are fewer atoms of B than required to react with all of A.

Solution approach

Correct Answer + Core Reason: A, 50 atoms; B, 30 atoms. B is the limiting reagent because there are fewer atoms of B than A, and the reaction requires a 1:1 ratio.

Step-by-Step Solution:

Step 1: The reaction is $A + B \to AB$ . This means 1 atom of A reacts with 1 atom of B.

Step 2: If B is the limiting reagent, it means there is not enough B to react with all of A.

Step 3: In option C, A has 50 atoms and B has 30 atoms. Since the ratio is 1:1, only 30 atoms of A will react with 30 atoms of B, leaving 20 atoms of A unreacted. Therefore, B is the limiting reagent.

Step 4: Final Answer: A, 50 atoms; B, 30 atoms

Why other options are wrong:
- Option A (A, 50 atoms; B, 100 atoms): B is not the limiting reagent because there are more atoms of B than A.
- Option B (A, 100 atoms; B, 6200 atoms): B is not the limiting reagent because there are significantly more atoms of B than A.
- Option D (A, 50 atoms; B, 200 atoms): B is not the limiting reagent because there are more atoms of B than A.

Key Concepts:
- Limiting reagent
- Stoichiometry

Memory Aids:
- The limiting reagent is the reactant that is completely consumed in a reaction.

Quick Tips:
- Compare the mole ratio of reactants to the stoichiometric ratio in the balanced equation.
- The reactant with the smallest mole ratio is the limiting reagent.

25 Medium

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Application +4 | -1

1 mole of chlorine combines with a certain weight of a metal giving 111 g of its chloride. The atomic weight of the metal (assuming its valency to be 2) is :

View answer and explanation

Correct option: C

80

Option C (80) is correct because the equivalent weight of the metal is 40, and since the valency is 2, the atomic weight is 80.

Solution approach

Correct Answer + Core Reason: 80. The atomic weight of the metal is calculated using the equivalent weight and valency.

Step-by-Step Solution:

Step 1: Write the equation for equivalents: $\mathrm{Eq}\mathrm{. ofC}{\mathrm{l}_2} = \mathrm{eq}\mathrm{. of chloride}$

Step 2: Substitute the given values: $1 \times 2 = \frac{{111}}{{\mathrm{E} + 35.5}}$

Step 3: Solve for E: $2E + 71 = 111$ , $2E = 40$ , $E = 20$

Step 4: Calculate the atomic weight: $M = E \times \text{valency} = 40 \times 2 = 80$

Why other options are wrong:
- Option A: 40 is incorrect because it is the equivalent weight, not the atomic weight. The atomic weight is twice the equivalent weight since the valency is 2.
- Option B: 20 is incorrect because it is half the equivalent weight, not the atomic weight.
- Option D: None of these is incorrect because 80 is the correct atomic weight.

Key Concepts:
- Equivalent weight
- Atomic weight
- Valency

Memory Aids:
- Atomic weight = Equivalent weight x Valency

Quick Tips:
- Use the relationship between equivalent weight, atomic weight, and valency.
- Ensure the units are consistent.

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Application +4 | -1

The molarity of

4.9 \%

(w / w)

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

solution having density

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

will be

View answer and explanation

Correct option: A

0.99 \mathrm{~M}

The correct answer is A (0.99 M) because it is calculated using the formula Molarity = (% by weight * density * 10) / Molar mass.

Solution approach

Correct Answer + Core Reason: 0.99 M. The molarity is calculated using the given percentage, density, and molar mass.

Step-by-Step Solution:

Step 1: Identify the given values.
% (w/w) of H2SO4 = 4.9%\nDensity = 1.98 g/mL\nMolar mass of H2SO4 = 98 g/mol

Step 2: Use the formula for molarity.\nMolarity (M) = (% by weight of solute * density * 10) / Molar mass of solute

Step 3: Substitute the values into the formula.\nM = (4.9 * 1.98 * 10) / 98

Step 4: Calculate the molarity.\nM = (4.9 * 1.98 * 10) / 98 = 97.02 / 98 = 0.990 M

Step 5: Round to two decimal places.\nM = 0.99 M

Why other options are wrong:
- Option B: 1.25 M - This is incorrect because it likely involves an error in the calculation or incorrect substitution in the formula.
- Option C: 0.12 M - This is incorrect because it likely involves an error in the calculation or incorrect substitution in the formula.
- Option D: 1.85 M - This is incorrect because it likely involves an error in the calculation or incorrect substitution in the formula.

Key Concepts:
- Molarity
- Percentage by weight

Memory Aids:
- Remember the formula: M = (% * d * 10) / Molar mass

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

27 Medium

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Application +4 | -1

How many moles of

\mathrm{P}_4 \mathrm{O}_6

and

\mathrm{P}_4 \mathrm{O}_{10}

will be produced by the combustion of

12.4 \mathrm{~g}

of phosphorous (atomic mass 31) in

12.8 \mathrm{~g}

of oxygen, leaving no

\mathrm{P}_4

or

\mathrm{O}_2

?

View answer and explanation

Correct option: C

0.05 \mathrm{~mol}

each

Option C (

0.05 \mathrm{~mol}

each) is correct because the reaction stoichiometry and limiting reactant determine the moles of

\mathrm{P}_4 \mathrm{O}_6

and

\mathrm{P}_4 \mathrm{O}_{10}

produced.

Solution approach

Correct Answer + Core Reason: 0.05 mol each. The limiting reactant and stoichiometry determine the product amounts.

Step-by-Step Solution:

Step 1: Calculate moles of $\mathrm{P}_4$ and $\mathrm{O}_2$ .\nMoles of $\mathrm{P}_4 = \frac{12.4}{4 \times 31} = 0.1 \mathrm{~mol}$ \nMoles of $\mathrm{O}_2 = \frac{12.8}{32} = 0.4 \mathrm{~mol}$

Step 2: Define the reactions and set up equations.\nLet $x$ moles of $\mathrm{P}_4$ form $\mathrm{P}_4 \mathrm{O}_6$ and $(0.1 - x)$ moles form $\mathrm{P}_4 \mathrm{O}_{10}$ .
P_4 + 3O_2 → P_4 O_6 $\n\mathrm{P}_4 + 5\mathrm{O}_2 \rightarrow \mathrm{P}_4 \mathrm{O}_{10}$

Step 3: Use the given information to solve for $x$ .\nMoles of $\mathrm{O}_2$ required $= 3x + 5(0.1 - x) = 0.4$
$3x + 0.5 - 5x = 0.4$
$-2x = -0.1$
$x = 0.05 \mathrm{~mol}$

Step 4: Calculate the moles of $\mathrm{P}_4 \mathrm{O}_6$ and $\mathrm{P}_4 \mathrm{O}_{10}$ .\nMoles of $\mathrm{P}_4 \mathrm{O}_6 = x = 0.05 \mathrm{~mol}$ \nMoles of $\mathrm{P}_4 \mathrm{O}_{10} = 0.1 - x = 0.1 - 0.05 = 0.05 \mathrm{~mol}$

Final Answer: 0.05 mol each

Why other options are wrong:
- Option A (0.1 and 0.3 mol): This is incorrect because it doesn't satisfy the mole balance of oxygen and phosphorus.
- Option B (0.15 mol and 0.25 mol): This is incorrect because it doesn't satisfy the mole balance of oxygen and phosphorus.
- Option D (0.1 mol each): This is incorrect because it doesn't satisfy the mole balance of oxygen and phosphorus and doesn't account for the limiting reactant.

Key Concepts:
- Stoichiometry
- Limiting Reactant
- Mole Concept

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

Quick Tips:
- Always identify the limiting reactant before calculating the amount of product formed.

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Application +4 | -1

How much

{AgCl}

will be formed by adding

200 \mathrm{~mL}

of

5 {~N}

{HCl}

to a solution containing

1.7 \mathrm{~g} \, {AgNO}_3

({Ag}=108)?

View answer and explanation

Correct option: B

1.435 \mathrm{~g}

Option B (1.435 g) is correct because it is calculated based on the limiting reagent and the stoichiometry of the reaction.

Solution approach

Correct Answer + Core Reason: 1.435 g. The amount of AgCl formed is determined by the limiting reagent.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $AgNO_3 + HCl \rightarrow AgCl + HNO_3$

Step 2: Calculate the number of moles of $AgNO_3$ : Moles of $AgNO_3 = \frac{{1.7 \text{{ g}}}}{{170 \text{{ g/mol}}}} = 0.01 \text{{ mol}}$ (Molar mass of $AgNO_3$ = 170 g/mol)

Step 3: Calculate the number of moles of $HCl$ : Given 200 mL of 5 N HCl. Normality = Molarity x n-factor. Since HCl is a monoprotic acid, n-factor = 1. So, Molarity = 5 N = 5 M.\nMoles of $HCl = Molarity \times Volume (in L) = 5 \times 0.2 = 1 \text{{ mol}}$

Step 4: Identify the limiting reagent: Since 1 mole of $AgNO_3$ reacts with 1 mole of $HCl$ , and we have 0.01 mol of $AgNO_3$ and 1 mol of $HCl$ , $AgNO_3$ is the limiting reagent.

Step 5: Calculate the number of moles of $AgCl$ formed: Since the stoichiometry is 1:1, 0.01 mol of $AgCl$ will be formed.

Step 6: Calculate the mass of $AgCl$ formed: Mass of $AgCl = Moles \times Molar mass = 0.01 \times 143.5 = 1.435 \text{{ g}}$ (Molar mass of $AgCl$ = 143.5 g/mol)

Why other options are wrong:
- Option A (0.1435 g): This is incorrect because it likely involves a calculation error or an incorrect application of the formula.
- Option C (14.35 g): This is incorrect because it likely involves a calculation error or an incorrect application of the formula.
- Option D (143.5 g): This is incorrect because it likely assumes that all of the HCl reacts, which is not the case since $AgNO_3$ is the limiting reagent.

Key Concepts:
- Stoichiometry
- Limiting reagent
- Mole concept
- Molar mass
- Normality and Molarity

Memory Aids:
- Remember to identify the limiting reagent before calculating the amount of product formed.

Quick Tips:
- Always write the balanced chemical equation first.
- Convert all given quantities to moles.
- Use the stoichiometry of the reaction to determine the amount of product formed.

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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.

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Remembering +4 | -1

One gram mole of a gas at NTP occupies 22.4 L. This fact was derived from

View answer and explanation

Correct option: B

Avogadro’s hypothesis

Option B (Avogadro’s hypothesis) is correct because Avogadro's hypothesis states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules, which leads to the molar volume concept.

Solution approach

Correct Answer + Core Reason: Avogadro’s hypothesis. Avogadro's hypothesis directly relates the volume of a gas to the number of moles.

Step-by-Step Solution:

Step 1: State Avogadro's hypothesis. Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

Step 2: Relate Avogadro's hypothesis to molar volume. This implies that one mole of any gas at NTP occupies the same volume.

Step 3: The volume occupied by one mole of a gas at NTP is 22.4 L.

Why other options are wrong:
- Option A: Law of gaseous volumes deals with the ratios of volumes of gases in chemical reactions, not directly with molar volume.
- Option C: Berzelius hypothesis (which was later proven incorrect) stated that equal volumes of all gases contain the same number of atoms, not molecules.
- Option D: Dalton’s atomic theory deals with the fundamental nature of atoms and their combinations, not directly with molar volume.

Key Concepts:
- Avogadro's hypothesis
- Molar volume
- NTP (Normal Temperature and Pressure)

Memory Aids:
- Avogadro = Volume relationship

Quick Tips:
- Remember that Avogadro's number (6.022 × 10²³) is related to the number of particles in one mole.

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Some basic concepts of chemistry practice · Medium

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