SOME BASIC CONCEPTS OF CHEMISTRY MCQs for JEE CHEMISTRY — Medium, Page 4

91 Medium

91 of 272

Application +4 | -1

An ethanolic solution of naphthalene contains 0.25 mole fraction of solute. Molality of solution is

View answer and explanation

Correct option: C

7.25 m

Option C (7.25 m) is correct because it is the molality of the solution calculated using the mole fraction of the solute and the molar mass of the solvent.

Solution approach

Correct Answer + Core Reason: 7.25 m. The molality of the solution is calculated using the mole fraction of the solute and the molar mass of the solvent (ethanol).

Step-by-Step Solution:

Step 1: Given: Mole fraction of solute (naphthalene), $\chi_A = 0.25$ . Therefore, mole fraction of solvent (ethanol), $\chi_B = 1 - 0.25 = 0.75$ .

Step 2: Molar mass of ethanol ( $C_2H_5OH$ ) is $2(12) + 6(1) + 16 = 24 + 6 + 16 = 46 \text{{ g/mol}}$ .

Step 3: Calculate molality (m) using the formula: $m = \frac{\chi_A}{\chi_B} \times \frac{1000}{M_B}$ , where $M_B$ is the molar mass of the solvent.

Step 4: Substitute the values: $m = \frac{0.25}{0.75} \times \frac{1000}{46} = \frac{1}{3} \times \frac{1000}{46} = \frac{1000}{138} = 7.246 \approx 7.25 \text{{ m}}$ .

Step 5: Final Answer: The molality of the solution is approximately 7.25 m.

Why other options are wrong:
- Option A (5.92 m): This is incorrect due to a miscalculation in the molality formula or incorrect substitution of values.
- Option B (3.39 m): This is incorrect due to a miscalculation in the molality formula or incorrect substitution of values.
- Option D (9.47 m): This is incorrect due to a miscalculation in the molality formula or incorrect substitution of values.

Key Concepts:
- Mole fraction
- Molality

Memory Aids:
- Molality is moles of solute per kilogram of solvent.

Quick Tips:
- Ensure you use the correct molar mass for the solvent.
- Remember the formula for molality when given mole fractions.

92 Medium

92 of 272

Application +4 | -1

Element

\mathrm{X}

reacts with oxygen to produce a pure sample of

\mathrm{X}_2 \mathrm{O}_3.

In an experiment it is found that

1 \mathrm{~g}

of

\mathrm{X}

produces

1.16 \mathrm{~g}

of

\mathrm{X}_2 \mathrm{O}_3.

Calculate the atomic weight of

\mathrm{X}.

(Given atomic weight of oxygen,

16.0 \mathrm{~g} \mathrm{~mol}^{-1}

)

View answer and explanation

Correct option: D

150

Option D (150) is correct because it is calculated using the principle of atom conservation (POAC) applied to element X.

Solution approach

Correct Answer + Core Reason: 150. The atomic weight is calculated using the principle of atom conservation.

Step-by-Step Solution:

Step 1: Write the reaction: $\mathrm{X} \rightarrow \mathrm{X}_2 \mathrm{O}_3$

Step 2: Apply POAC on X: $1 \times (\text{No. of moles of } \mathrm{X}) = 2 \times (\text{No. of moles of } \mathrm{X}_2 \mathrm{O}_3)$

Step 3: Let the atomic weight of X be 'a'.

Step 4: Substitute the given values: $1 \times \frac{1}{a} = 2 \times \frac{1.16}{(2a + 3 \times 16)}$

Step 5: Simplify the equation: $\frac{1}{a} = \frac{2.32}{2a + 48}$

Step 6: Solve for 'a': $2a + 48 = 2.32a$ , so $0.32a = 48$ , and $a = \frac{48}{0.32} = 150$

Why other options are wrong:
- Option A (67): This is incorrect because of an error in applying POAC or in the calculation.
- Option B (100.2): This is incorrect because of an error in applying POAC or in the calculation.
- Option C (125): This is incorrect because of an error in applying POAC or in the calculation.

Key Concepts:
- Principle of Atom Conservation (POAC)
- Stoichiometry

Memory Aids:
- Remember that POAC states that the number of atoms of an element remains constant in a chemical reaction.

Quick Tips:
- Always write the balanced chemical equation (even if it's just a transformation) before applying POAC.
- Make sure to account for the number of atoms of the element in each molecule.

93 Medium

93 of 272

Understanding +4 | -1

Potassium combines with two isotopes of chlorine

\left({ }^{35} {Cl} \text{ and } { }^{37} {Cl}\right)

respectively to form two samples of

{KCl}.

Their formation follows the law of

View answer and explanation

Correct option: A

Constant proportions

Option A (Constant proportions) is correct because the formation of KCl from potassium and either isotope of chlorine always follows the law of constant proportions.

Solution approach

Correct Answer + Core Reason: Constant proportions. The law of constant proportions states that a given chemical compound always contains its constituent elements in a fixed ratio (by mass) and does not depend on its source and method of preparation.

Step-by-Step Solution:

Step 1: Recall the Law of Constant Proportions (also known as the Law of Definite Proportions).

Step 2: Consider the formation of KCl from potassium and chlorine isotopes.
- Potassium (K) combines with chlorine isotopes ${ }^{35}Cl$ and ${ }^{37}Cl$ to form KCl.
- Regardless of which chlorine isotope is used, the ratio of potassium to chlorine in KCl will always be constant.

Step 3: Apply the Law of Constant Proportions.
- The formation of KCl follows the Law of Constant Proportions because the ratio of potassium to chlorine is fixed in the compound.

Why other options are wrong:
- Option B: Incorrect because the Law of Multiple Proportions applies when two elements form more than one compound, which is not the case here.
- Option C: Incorrect because the Law of Reciprocal Proportions applies when three or more elements combine in different ways, which is not the case here.
- Option D: Incorrect because the formation of KCl does follow a law of chemical combination.

Key Concepts:
- Law of Constant Proportions
- Law of Multiple Proportions
- Law of Reciprocal Proportions

Memory Aids:
- Constant Proportions: A compound always has the same proportion of elements by mass.

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

94 Medium

94 of 272

Analyzing +4 | -1

Assertion (A): The percentage of nitrogen in urea is 46.6%.

Reason (R): Urea is a covalent compound.

View answer and explanation

Correct option: B

Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

Option B is correct. Both the assertion and the reason are true, but the reason is not the correct explanation for the assertion. The percentage of nitrogen in urea is 46.6%, and urea is a covalent compound, but the covalent nature doesn't explain the nitrogen percentage.

Solution approach

Correct Answer + Core Reason: Both (A) & (R) are true but the (R) is not the correct explanation of the (A). The percentage of nitrogen in urea is a result of its molecular formula, not just its covalent nature.

Step-by-Step Solution:

Step 1: Calculate the percentage of nitrogen in urea ( $\mathrm{CO(NH_2)_2}$ ). The molar mass of urea is $12 + 16 + 2(14 + 2(1)) = 60 \text{{ g/mol}}$ .

Step 2: Calculate the mass of nitrogen in one mole of urea: $2 \times 14 = 28 \text{{ g}}$ .

Step 3: Calculate the percentage of nitrogen: $\frac{28}{60} \times 100 \approx 46.6\%$ .

Step 4: Evaluate the Reason: Urea is indeed a covalent compound.

Step 5: Determine if the reason explains the assertion: The reason doesn't explain why the percentage of nitrogen is 46.6%. The percentage is due to the ratio of nitrogen atoms to the total mass of the molecule.

Why other options are wrong:
- Option A: The reason is not the correct explanation.
- Option C: The reason is true, so this option is incorrect.
- Option D: Both the assertion and reason are true, so this option is incorrect.

Key Concepts:
- Percentage Composition
- Covalent Compounds

Memory Aids:
- Remember the formula for calculating percentage composition: (mass of element / molar mass of compound) * 100.

Quick Tips:
- Calculate the molar mass of the compound and the mass of the element of interest to determine the percentage composition.

95 Medium

95 of 272

Understanding +4 | -1

AgN{O_3} + KI \to AgI + KN{O_3}

. This reaction is example for.

View answer and explanation

Correct option: A

Law of conservation of mass

The correct answer is A (Law of conservation of mass) because the reaction demonstrates that the total mass of reactants equals the total mass of products.

Solution approach

Correct Answer + Core Reason: Law of conservation of mass. The reaction exemplifies the conservation of mass principle.

Step-by-Step Solution:

Step 1: Identify the reaction: $AgNO_3 + KI \to AgI + KNO_3$

Step 2: Key Concept: Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction.

Step 3: Explanation: The given reaction shows a chemical change where the total mass of reactants ( $AgNO_3$ and $KI$ ) is equal to the total mass of products ( $AgI$ and $KNO_3$ ).

Step 4: Conclusion: This reaction exemplifies the Law of Conservation of Mass.

Why other options are wrong:
- Option B: Law of definite proportion states that a chemical compound always contains exactly the same proportion of elements by mass. This reaction doesn't illustrate that.
- Option C: Law of multiple proportion states that if two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers. This reaction doesn't illustrate that.
- Option D: Law of reciprocal proportion states that if two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the ratio of the masses in which they combine with each other. This reaction doesn't illustrate that.

Key Concepts:
- Law of Conservation of Mass
- Chemical Reactions

Memory Aids:
- Conservation: "Conserve" means to keep the same.

Quick Tips:
- Remember the definitions of each law to identify the correct one in different scenarios.

96 Medium

96 of 272

Application +4 | -1

Find the weight of

\mathrm{KOH}

in its 50 milli equivalents

View answer and explanation

Correct option: C

2.8

Option C (2.8) is correct because it is the calculated weight of KOH based on its milli-equivalent weight.

Solution approach

Correct Answer + Core Reason: 2.8. The weight of KOH is calculated using the definition of milli-equivalents.

Step-by-Step Solution:

Step 1: Define milli-equivalents (meq): $\text{meq} = \frac{{\text{weight}}}{{\text{Equivalent weight}}} \times 1000$

Step 2: Calculate the equivalent weight of KOH: Equivalent weight = $\frac{{\text{Molecular weight}}}{{\text{Basicity}}} = \frac{{56}}{{1}} = 56$

Step 3: Substitute the given values into the formula: $50 = \frac{{\text{weight}}}{{56}} \times 1000$

Step 4: Solve for the weight of KOH: $\text{weight} = \frac{{50 \times 56}}{{1000}} = \frac{{2800}}{{1000}} = 2.8 \text{ g}$

Why other options are wrong:
- Option A (1.6): Incorrect. This value likely arises from an error in the calculation, possibly using an incorrect equivalent weight or misapplying the formula for milli-equivalents.
- Option B (2.2): Incorrect. This value likely arises from an error in the calculation, possibly using an incorrect equivalent weight or misapplying the formula for milli-equivalents.
- Option D (4.8): Incorrect. This value likely arises from an error in the calculation, possibly using an incorrect equivalent weight or misapplying the formula for milli-equivalents.

Key Concepts:
- Equivalent weight
- Milli-equivalents
- Acid-base stoichiometry

Memory Aids:
- Remember the formula: $\text{meq} = \frac{{\text{weight}}}{{\text{Equivalent weight}}} \times 1000$

Quick Tips:
- Always calculate the equivalent weight correctly based on the acidity or basicity of the substance.
- Pay attention to the units and convert them if necessary.

97 Medium

97 of 272

Application +4 | -1

Number of g of oxygen in 32.2 g

Na_2SO_4.10H_2O

is

View answer and explanation

Correct option: C

22.4

Option C (22.4) is correct because the mass of oxygen is calculated based on the molar mass of the compound and the number of oxygen atoms present.

Solution approach

Correct Answer + Core Reason: 22.4. The mass of oxygen is calculated based on the molar mass of the compound and the number of oxygen atoms present.

Step-by-Step Solution:

Step 1: Determine the molar mass of $Na_2SO_4 \cdot 10H_2O$ : $2(23) + 32 + 4(16) + 10[2(1) + 16]

98 Medium

98 of 272

Application +4 | -1

The number of electrons in

0.2 \mathrm{~mol}

of

{Al}^{3+}

ion is

View answer and explanation

Correct option: D

2 \times \mathrm{N}_{\mathrm{A}}

Option D (

2 \times N_A

) is correct because each Al3+ ion has 10 electrons, and 0.2 moles of Al3+ ions will therefore have 2NA electrons.

Solution approach

Correct Answer + Core Reason: $2 \times N_A$ . The total number of electrons is calculated by multiplying the number of moles, Avogadro's number, and the number of electrons per ion.

Step-by-Step Solution:

Step 1: Determine the number of electrons in a neutral Aluminum (Al) atom: 13 electrons.

Step 2: Determine the number of electrons in an $Al^{3+}$ ion: Since it lost 3 electrons, $13 - 3 = 10$ electrons.

Step 3: Calculate the total number of electrons in 0.2 moles of $Al^{3+}$ ions: $0.2 \, mol \times 10 \, electrons/ion \times N_A \, electrons/mol = 2N_A$ electrons.

Why other options are wrong:
- **Option A ( $0.2 \times N_A$ ):** This is incorrect because it only considers 0.2 moles of electrons without accounting for the 10 electrons in each $Al^{3+}$ ion.
- **Option B ( $0.1 \times N_A$ ): This is incorrect as it doesn't correctly account for the number of electrons in each ion and the number of moles.
- Option C ( $N_A$ ):** This is incorrect because it assumes only 1 mole of electrons, without considering the 0.2 moles of $Al^{3+}$ ions and the 10 electrons per ion.

Key Concepts:
- Mole concept
- Avogadro's number ( $N_A$ )
- Electronic configuration of ions

Memory Aids:
- Remember that a positive charge on an ion indicates the loss of electrons.

Quick Tips:
- Always consider the charge of the ion when calculating the number of electrons.
- Make sure to multiply by Avogadro's number when converting moles to the number of particles.

99 Medium

99 of 272

Understanding +4 | -1

Assertion : One molal aqueous solution of glucose contains

180 {~g}

of glucose in

1 {~kg}

water.

Reason : Solution containing one mole of solute in

1000 {~g}

of solvent is called one molal solution.

View answer and explanation

Correct option: A

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

Option A (If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion) is correct because the definition of molality directly explains why a 1 molal glucose solution contains 180g of glucose in 1 kg of water.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion. The definition of molality supports the assertion.

Step-by-Step Solution:

Step 1: Define molality: Molality is defined as the number of moles of solute per kilogram of solvent.

Step 2: Consider the assertion: One molal aqueous solution of glucose contains 180 g of glucose in 1 kg water.

Step 3: Calculate the number of moles of glucose in 180 g: Molar mass of glucose ( $C_6H_{12}O_6$ ) = 180 g/mol. Therefore, 180 g of glucose is equal to 1 mole.

Step 4: Relate the assertion to the definition of molality: A one molal solution contains 1 mole of solute in 1 kg of solvent. Since the solution contains 1 mole of glucose in 1 kg of water, it is a one molal solution.

Step 5: The reason states: Solution containing one mole of solute in 1000 g of solvent is called one molal solution. This is the correct definition of molality.

Step 6: The reason correctly explains the assertion because the definition of molality directly supports the statement that a one molal glucose solution contains 180 g of glucose in 1 kg of water.

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

Key Concepts:
- Molality
- Molar Mass
- Solutions
- Concentration

Memory Aids:
- Molality = moles of solute / kg of solvent.

Quick Tips:
- Remember the definition of molality and how to calculate moles.

100 Medium

100 of 272

Application +4 | -1

The total number of electrons present in

18 \mathrm{~mL}

of water (density of water is

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

) is

View answer and explanation

Correct option: C

6.02 \times 10^{24}

Option C (

6.02 \times 10^{24}

) is correct because 18 mL of water is 1 mole, and each water molecule has 10 electrons.

Solution approach

Correct Answer + Core Reason: $6.02 \times 10^{24}$ . 18 mL of water is 1 mole, and each water molecule has 10 electrons.

Step-by-Step Solution:

Step 1: Given: Volume of water = 18 mL, Density of water = 1 g/mL

Step 2: Calculate the mass of water: Mass = Density x Volume = 1 g/mL x 18 mL = 18 g

Step 3: Calculate the number of moles of water: Moles = Mass / Molar mass = 18 g / 18 g/mol = 1 mol

Step 4: Calculate the number of water molecules: Number of molecules = Moles x Avogadro's number = 1 mol x $6.022 \times 10^{23}$ molecules/mol = $6.022 \times 10^{23}$ molecules

Step 5: Determine the number of electrons in one water molecule: Each hydrogen atom has 1 electron, and the oxygen atom has 8 electrons. Therefore, $H_2O$ has (2 x 1) + 8 = 10 electrons.

Step 6: Calculate the total number of electrons: Total electrons = Number of molecules x Electrons per molecule = $6.022 \times 10^{23}$ molecules x 10 electrons/molecule = $6.022 \times 10^{24}$ electrons

Step 7: Final Answer: $6.02 \times 10^{24}$

Why other options are wrong:
- **Option A ( $6.02 \times 10^{23}$ ): This is incorrect because it only calculates the number of water molecules, not the total number of electrons.
- Option B ( $6.02 \times 10^{22}$ ): This is incorrect because it has an incorrect exponent.
- Option D ( $6.02 \times 10^{25}$ ): This is incorrect because it has an incorrect exponent.

Key Concepts:
- Mole Concept
- Avogadro's Number
- Molar Mass

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

Quick Tips:
- Always convert mass to moles before calculating the number of particles.
- Double-check the number of electrons in each atom.

101 Medium

101 of 272

Application +4 | -1

0.1 gram mole of urea is dissolved in 100 g of water. The molarity of the solution is.

View answer and explanation

Correct option: D

1.0 M

The correct answer is D (1.0 M) because the molarity is calculated by dividing the moles of solute by the volume of the solution in liters.

Solution approach

Correct Answer + Core Reason: 1.0 M. The molarity is calculated using the formula: Molarity = moles of solute / volume of solution (in liters).

Step-by-Step Solution:

Step 1: Identify the given information: Moles of urea = 0.1 gram mole, mass of water = 100 g.

Step 2: Assume the density of the solution is approximately equal to the density of water (1 g/mL). Therefore, the volume of the solution is approximately 100 mL.

Step 3: Convert the volume of the solution to liters: 100 mL = 0.1 L.

Step 4: Calculate the molarity: Molarity = $\frac{\text{{moles of urea}}}{\text{{volume of solution in liters}}} = \frac{0.1 \text{{ moles}}}{0.1 \text{{ L}}}$

Step 5: Molarity = 1.0 M

Why other options are wrong:
- Option A (1 M): Correct answer.
- Option B (0.02 M): Incorrect calculation.
- Option C (0.01 M): Incorrect calculation.

Key Concepts:
- Molarity
- Density of water

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

Quick Tips:
- Remember to convert the volume to liters before calculating molarity.
- If the density of the solution is not given, assume it is approximately equal to the density of water for dilute solutions.

102 Medium

102 of 272

Application +4 | -1

Volume of

C{O_2}

obtained by the complete decomposition of 9.85 g of

BaC{O_3}

is (at. wt of Ba = 137)

View answer and explanation

Correct option: B

1.12 L

Option B (1.12 L) is correct because it is the volume of CO2 produced from the complete decomposition of 9.85 g of BaCO3, calculated using stoichiometry.

Solution approach

Correct Answer + Core Reason: 1.12 L. The volume of CO2 is calculated using stoichiometry from the balanced equation for the decomposition of BaCO3.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $BaCO_3(s) \rightarrow BaO(s) + CO_2(g)$

Step 2: Calculate the molar mass of $BaCO_3$ : $137 + 12 + (3 \times 16) = 197 \text{{ g/mol}}$

Step 3: Calculate the number of moles of $BaCO_3$ : $\frac{{9.85 \text{{ g}}}}{{197 \text{{ g/mol}}}} = 0.05 \text{{ mol}}$

Step 4: From the balanced equation, 1 mole of $BaCO_3$ produces 1 mole of $CO_2$ . Therefore, 0.05 moles of $BaCO_3$ will produce 0.05 moles of $CO_2$ .

Step 5: Calculate the volume of $CO_2$ at standard conditions (assuming STP): $V = nRT/P$ . At STP, T = 273 K and P = 1 atm. Using R = 0.0821 L atm / (mol K), $V = (0.05 \text{{ mol}}) \times (0.0821 \text{{ L atm / (mol K)}}) \times (273 \text{{ K}}) / (1 \text{{ atm}}) = 1.12 \text{{ L}}$

Why other options are wrong:
- Option A (2.4 L): This is incorrect as it likely involves an error in the stoichiometric calculation or incorrect use of the ideal gas law.
- Option C (0.84 L): This is incorrect and could arise from using an incorrect molar mass or a mistake in the mole ratio.
- Option D (0.56 L): This is incorrect and may be due to an error in the calculation of moles or volume using the ideal gas law.

Key Concepts:
- Stoichiometry
- Molar mass calculation
- Mole concept
- Ideal gas law (at STP)

Memory Aids:
- Remember the relationship between moles, mass, and molar mass: moles = mass / molar mass
- At STP, 1 mole of any gas occupies 22.4 L (approximately), but using the ideal gas law is more accurate when not exactly at STP.

Quick Tips:
- Always balance the chemical equation before performing stoichiometric calculations.
- Ensure correct units are used in the ideal gas law equation.

103 Medium

103 of 272

Application +4 | -1

x\;g\;of\;Ag

was dissolved in

\mathrm{HN}{\mathrm{O}_3}

and the solution was treated with excess of

NaCl

when 2.87 g of

AgCl

was precipitated. The value of x is

View answer and explanation

Correct option: B

2.16\mathrm{ }\;g

Option B (2.16 g) is correct because it is calculated using stoichiometry based on the amount of AgCl precipitated.

Solution approach

Correct Answer + Core Reason: 2.16 g. The mass of Ag is determined using stoichiometry based on the mass of AgCl.

Step-by-Step Solution:

Step 1: Write the relevant reaction: $Ag^+ + Cl^- \rightarrow AgCl$

Step 2: Establish the molar relationship: 1 mole of AgCl corresponds to 1 mole of Ag.

Step 3: Calculate the molar mass of AgCl: $107.87 + 35.45 = 143.32 g/mol$ (approximately 143.5 g/mol).

Step 4: Calculate the moles of AgCl precipitated: $\frac{{2.87 g}}{{143.5 g/mol}} = 0.02 mol$

Step 5: Since moles of Ag = moles of AgCl, moles of Ag = 0.02 mol.

Step 6: Calculate the mass of Ag: $0.02 mol \times 107.87 g/mol = 2.1574 g$ (approximately 2.16 g).

Step 7: Final Answer: 2.16 g

Why other options are wrong:
- Option A (1.08 g): This is incorrect. It might be obtained by dividing the mass of AgCl by 2 instead of using the correct stoichiometric ratio.
- Option C (2.70 g): This is incorrect. This value doesn't align with the correct stoichiometric calculation.
- Option D (1.62 g): This is incorrect. This value doesn't align with the correct stoichiometric calculation.

Key Concepts:
- Stoichiometry
- Mole concept
- Precipitation reactions

Memory Aids:
- Remember to use the correct molar masses.

Quick Tips:
- Always write the balanced chemical equation.
- Pay attention to the stoichiometric ratios.

104 Medium

104 of 272

Analyzing +4 | -1

Which of the following has the smallest number of molecules?

View answer and explanation

Correct option: A

0.1 mole of

\mathrm{C}{\mathrm{O}_2}

gas

Option A (0.1 mole of

\mathrm{C}{\mathrm{O}_2}

gas) is correct because it represents the smallest number of moles, and equal numbers of moles have equal numbers of molecules.

Solution approach

Correct Answer + Core Reason: 0.1 mole of $\mathrm{C}{\mathrm{O}_2}$ gas. The smallest number of molecules corresponds to the smallest number of moles.

Step-by-Step Solution:

Step 1: Convert all options to moles of $\mathrm{C}{\mathrm{O}_2}$ .

Step 2: Option A: 0.1 mole of $\mathrm{C}{\mathrm{O}_2}$ gas.

Step 3: Option B: 11.2 L of $\mathrm{C}{\mathrm{O}_2}$ gas at STP. Using the ideal gas law at STP, 1 mole occupies 22.4 L. Therefore, moles = $\frac{{11.2}}{{22.4}} = 0.5$ mole.

Step 4: Option C: 22 g of $\mathrm{C}{\mathrm{O}_2}$ gas. The molar mass of $\mathrm{C}{\mathrm{O}_2}$ is 44 g/mol. Therefore, moles = $\frac{{22}}{{44}} = 0.5$ mole.

Step 5: Option D: $22.4 \times {10^3}$ mL of $\mathrm{C}{\mathrm{O}_2}$ gas at STP. Convert mL to L: $22.4 \times {10^3}$ mL = 22.4 L. Therefore, moles = $\frac{{22.4}}{{22.4}} = 1$ mole.

Step 6: Compare the number of moles for each option: A (0.1), B (0.5), C (0.5), D (1).

Step 7: The smallest number of moles is 0.1 mole.

Why other options are wrong:
- Option B: 11.2 L of $\mathrm{C}{\mathrm{O}_2}$ gas at STP corresponds to 0.5 moles, which is greater than 0.1 moles.
- Option C: 22 g of $\mathrm{C}{\mathrm{O}_2}$ gas corresponds to 0.5 moles, which is greater than 0.1 moles.
- Option D: $22.4 \times {10^3}$ mL of $\mathrm{C}{\mathrm{O}_2}$ gas at STP corresponds to 1 mole, which is greater than 0.1 moles.

Key Concepts:
- Mole concept
- Ideal gas law at STP (1 mole = 22.4 L)
- Molar mass

Memory Aids:
- STP: Standard Temperature and Pressure
- Molar mass: Mass of one mole of a substance

Quick Tips:
- Always convert to moles when comparing the amount of substance.
- Remember the molar volume of a gas at STP (22.4 L/mol).

105 Medium

105 of 272

Application +4 | -1

Number of g-atom of S present in

\mathrm{49}\;\mathrm{g}\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

are:

View answer and explanation

Correct option: A

0.5

Option A (0.5) is correct because 49 g of H2SO4 contains 0.5 g-atom of S, calculated based on the molar mass of H2SO4 and the atomic mass of S.

Solution approach

Correct Answer + Core Reason: 0.5. The number of g-atoms of S is calculated based on the molar mass of H2SO4.

Step-by-Step Solution:

Step 1: Determine the molar mass of H2SO4: $2(1) + 32 + 4(16) = 98 \text{{ g/mol}}$

Step 2: 98 g of H2SO4 contains 1 g-atom (1 mole) of S.

Step 3: Calculate the number of g-atoms of S in 49 g of H2SO4: $\frac{{49 \text{{ g}}}}{{98 \text{{ g/mol}}}} \times 1 \text{{ g-atom}} = 0.5 \text{{ g-atom}}$

Step 4: Final Answer: 0.5

Why other options are wrong:
- Option B (1): This is incorrect because it corresponds to 98g of H2SO4, not 49g.
- Option C (0.2): This is an incorrect calculation.
- Option D (0.3): This is an incorrect calculation.

Key Concepts:
- Gram-atom
- Molar mass
- Stoichiometry

Memory Aids:
- 1 mole of a substance contains 1 gram-atom of each element present in the substance.

Quick Tips:
- Always calculate the molar mass correctly.

106 Medium

106 of 272

Understanding +4 | -1

A: Mass of

1 \mathrm{gram}

molecule of

{H}_2{SO}_4

is

98 \mathrm{gram}

.

R: One gram atom contains

{N}_{{A}}

atoms.

View answer and explanation

Correct option: B

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

Option B (Both A and R are correct, but R is not the correct explanation of A) is correct because both statements are factually correct, but the reason doesn't explain the assertion.

Solution approach

Correct Answer + Core Reason: Both A and R are correct, but R is not the correct explanation of A. Both statements are true, but unrelated.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): Mass of 1 gram molecule of ${H}_2{SO}_4$ is 98 gram.

Step 2: The molar mass of ${H}_2{SO}_4$ is 2(1) + 32 + 4(16) = 98 g/mol.

Step 3: Therefore, Assertion (A) is correct.

Step 4: Analyze Reason (R): One gram atom contains ${N}_{{A}}$ atoms.

Step 5: One gram atom is equal to one mole of atoms, which contains Avogadro's number ( ${N}_{{A}}$ ) of atoms.

Step 6: Therefore, Reason (R) is correct.

Step 7: Reason (R) does not explain Assertion (A).

Why other options are wrong:
- Option A (Both A and R are correct and R is the correct explanation of A): R does not explain A.
- Option C (A is correct but R is incorrect): R is correct.
- Option D (Both the A and R are incorrect): Both A and R are correct.

Key Concepts:
- Molar mass
- Gram molecule
- Gram atom
- Avogadro's number

Memory Aids:
- 1 gram molecule = 1 mole
- 1 gram atom = 1 mole of atoms

Quick Tips:
- Understand the definitions of gram molecule and gram atom.

107 Medium

107 of 272

Analyzing +4 | -1

Assertion: On increasing the temperature, the millimoles of solute, milliequivalent of solute, molality, mole fraction of solute and

\%

by weight does not change.
Reason: Each of these involves only masses of solute and solvent.

View answer and explanation

Correct option: C

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

Option C is correct because both the assertion and reason are correct, and the reason accurately explains why the listed concentration units are independent of temperature.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct and Reason is correct explanation of Assertion. Temperature has no effect on mass.

Step-by-Step Solution:

Step 1: Analyze the assertion. The assertion states that millimoles of solute, milliequivalent of solute, molality, mole fraction of solute, and % by weight do not change with temperature.

Step 2: Analyze the reason. The reason states that each of these concentration units involves only masses of solute and solvent.

Step 3: Evaluate the relationship between the assertion and the reason. Since mass is independent of temperature, any concentration unit that is based solely on mass will also be independent of temperature. The listed concentration units (millimoles, milliequivalent, molality, mole fraction, and % by weight) are all defined using masses of solute and solvent.

Step 4: Conclude that the assertion is correct because the listed concentration units are based on mass, which is independent of temperature. The reason correctly explains why the assertion is true.

Why other options are wrong:
- Option A: The assertion is correct, so this option is wrong.
- Option B: The assertion is correct, so this option is wrong.
- Option D: The reason is the correct explanation for the assertion, so this option is wrong.

Key Concepts:
- Concentration units: molality, mole fraction, % by weight
- Temperature dependence of mass and volume

Memory Aids:
- Remember that mass is conserved and does not change with temperature.

Quick Tips:
- When dealing with concentration units, always consider whether they are based on mass or volume. Mass-based units are temperature-independent, while volume-based units are temperature-dependent.

108 Medium

108 of 272

Remembering +4 | -1

Law of constant composition is same as the law of

View answer and explanation

Correct option: D

Definite proportion

Option D (Definite proportion) is correct because the law of constant composition is another name for the law of definite proportions.

Solution approach

Correct Answer + Core Reason: Definite proportion. The law of constant composition is the same as the law of definite proportions.

Step-by-Step Solution:

Step 1: Recall the law of constant composition.

Step 2: The law of constant composition states that a given chemical compound always contains the same elements in the same proportion by mass.

Step 3: The law of definite proportions is another name for the law of constant composition.

Why other options are wrong:
- Option A: Conservation of mass states that mass is neither created nor destroyed in a chemical reaction.
- Option B: Conservation of energy states that energy is neither created nor destroyed in a chemical reaction.
- Option C: Multiple proportion 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.

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

Memory Aids:
- Remember that constant composition and definite proportions are synonymous.

Quick Tips:
- Know the different laws of chemical combination.

109 Medium

109 of 272

Application +4 | -1

An aqueous solution has molality 11.11. Hence mole fraction of solute in the solution is

View answer and explanation

Correct option: A

0.167

Option A (0.167) is correct because it is the calculated mole fraction of the solute in the aqueous solution with a molality of 11.11.

Solution approach

Correct Answer + Core Reason: 0.167. The mole fraction of the solute is calculated using the molality of the solution.

Step-by-Step Solution:

Step 1: Define molality: Molality is defined as the number of moles of solute per kilogram of solvent.

Step 2: Given molality = 11.11, this means there are 11.11 moles of solute in 1 kg of solvent (water).

Step 3: Convert the mass of water to moles: 1 kg of water = 1000 g of water. The molar mass of water is 18 g/mol. So, moles of water = $\frac{{1000}}{{18}} = 55.55 \mathrm{~mol}$ .

Step 4: Calculate the mole fraction of the solute: Mole fraction of solute = $\frac{{\text{{moles of solute}}}}{{\text{{moles of solute}} + \text{{moles of solvent}}}} = \frac{{11.11}}{{11.11 + 55.55}}$ .

Step 5: Simplify the expression: Mole fraction of solute = $\frac{{11.11}}{{66.66}} = 0.16665 \approx 0.167$ .

Step 6: Final Answer: 0.167.

Why other options are wrong:
- Option B: 0.833: This is the mole fraction of the solvent, not the solute.
- Option C: 0.4: This is incorrect due to a miscalculation or incorrect application of the formula.
- Option D: 0.6: This is incorrect due to a miscalculation or incorrect application of the formula.

Key Concepts:
- Molality
- Mole fraction
- Molar mass

Memory Aids:
- Molality (m) = moles of solute / kg of solvent
- Mole fraction (X) = moles of component / total moles

Quick Tips:
- Remember the definitions of molality and mole fraction.
- Ensure the units are consistent throughout the calculation.

110 Medium

110 of 272

Analyzing +4 | -1

Assertion (A): The percentage of nitrogen in urea is 46.6%.

Reason (R): Urea is Ionic compound.

View answer and explanation

Correct option: C

(A) is true but (R) is false

Option C ((A) is true but (R) is false) is correct because the percentage of nitrogen in urea is indeed 46.6%, but urea is a covalent, not ionic, compound.

Solution approach

Correct Answer + Core Reason: (A) is true but (R) is false. The percentage of nitrogen in urea is correct, but urea is not an ionic compound.

Step-by-Step Solution:

Step 1: Verify the percentage of nitrogen in urea (NH2CONH2).

Step 2: Calculate the molar mass of urea: 12 (C) + 4(1) (H) + 2(14) (N) + 16 (O) = 60 g/mol.

Step 3: Calculate the mass of nitrogen in urea: 2 * 14 = 28 g/mol.

Step 4: Calculate the percentage of nitrogen: (28/60) * 100 = 46.6%.

Step 5: Determine the bonding type in urea. Urea is a covalent compound due to the sharing of electrons between atoms.

Why other options are wrong:
- Option A: Incorrect because the reason is false.
- Option B: Incorrect because the reason is false.
- Option D: Incorrect because the assertion is true.

Key Concepts:
- Percentage composition
- Chemical bonding (covalent vs. ionic)

Memory Aids:
- Remember the formula for percentage composition: (mass of element / molar mass of compound) * 100

Quick Tips:
- Always double-check the bonding type of compounds.

111 Medium

111 of 272

Application +4 | -1

Percentage loss in mass, when

{NaHCO}_3(\mathrm{~s})

is heated in open vessel

2 {NaHCO}_3(\mathrm{~s}) \rightarrow {Na}_2 \mathrm{CO}_3(\mathrm{~s})+{CO}_2(\mathrm{~g})+{H}_2 {O}(\mathrm{g})

View answer and explanation

Correct option: B

36.9 \%

Option B (36.9%) is correct because the percentage loss in mass is calculated based on the molar mass of the gases released (CO₂ and H₂O) relative to the molar mass of NaHCO₃.

Solution approach

Correct Answer + Core Reason: 36.9 %. The percentage loss in mass is calculated based on the stoichiometry of the decomposition reaction.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation:
$2 {NaHCO}_3(\mathrm{~s}) \rightarrow {Na}_2 {CO}_3(\mathrm{~s})+{CO}_2(\mathrm{~g})+{H}_2 {O}(\mathrm{g})$

Step 2: Calculate the molar mass of ${NaHCO}_3$ : 23 + 1 + 12 + (3 * 16) = 84 g/mol

Step 3: Calculate the molar mass of ${CO}_2$ : 12 + (2 * 16) = 44 g/mol

Step 4: Calculate the molar mass of ${H}_2 {O}$ : (2 * 1) + 16 = 18 g/mol

Step 5: For every 2 moles of ${NaHCO}_3$ (2 * 84 = 168 g), 1 mole of ${CO}_2$ (44 g) and 1 mole of ${H}_2 {O}$ (18 g) are released.

Step 6: Calculate the total mass loss: 44 + 18 = 62 g

Step 7: Calculate the percentage weight loss:
$\text{{Percentage Weight loss}} = \frac{{\text{{Mass loss}}}}{{\text{{Initial mass}}}} \times 100 = \frac{{62}}{{168}} \times 100 = 36.9 \%$

Final Answer: 36.9 %

Why other options are wrong:
- Option A: Incorrect calculation of the mass loss or the initial mass.
- Option C: Incorrect calculation of the mass loss or the initial mass.
- Option D: Incorrect calculation of the mass loss or the initial mass.

Key Concepts:
- Stoichiometry
- Molar mass calculation
- Percentage composition

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

Quick Tips:
- Always identify the gases released in a decomposition reaction to calculate the mass loss correctly.

112 Medium

112 of 272

Application +4 | -1

1.5 g of a divalent metal displaced 4 g of copper (at. wt. = 63.8) from a solution of copper sulphate. The atomic weight of the metal is:

View answer and explanation

Correct option: B

24

Option B (24) is correct because it correctly calculates the atomic weight of the divalent metal using the concept of equivalent weights.

Solution approach

Correct Answer + Core Reason: 24. The atomic weight is calculated using the concept of equivalent weights, where the equivalents of the metal and copper are equal.

Step-by-Step Solution:

Step 1: State the given information: Mass of metal = 1.5 g, Mass of copper displaced = 4 g, Atomic weight of copper = 63.8, Valency of metal = 2.

Step 2: Apply the concept of equivalent weights: Equivalent of metal = Equivalent of copper.

Step 3: Write the formula for equivalent weight: Equivalent weight = $\frac{{\text{Atomic weight}}}{{\text{Valency}}}$

Step 4: Let E be the equivalent weight of the metal. Then, $\frac{{1.5}}{E} = \frac{4}{{\frac{{63.8}}{2}}}$

Step 5: Solve for E: $E = \frac{{1.5 \times \frac{{63.8}}{2}}}{4} = \frac{{1.5 \times 31.9}}{4} = \frac{{47.85}}{4} = 11.9625 \approx 12$

Step 6: Calculate the atomic weight of the metal: Atomic weight = Equivalent weight × Valency = 12 × 2 = 24

Why other options are wrong:
- Option A (12): This value is incorrect because it only calculates the equivalent weight and doesn't multiply by the valency to find the atomic weight.
- Option C (48): This value is incorrect and likely arises from multiplying the equivalent weight by a wrong factor or an arithmetic error.
- Option D (6): This value is incorrect and suggests a significant error in the calculation, possibly involving an incorrect division or multiplication.

Key Concepts:
- Equivalent weight
- Atomic weight
- Stoichiometry

Memory Aids:
- Remember that Equivalent weight = $\frac{{\text{Atomic weight}}}{{\text{Valency}}}$
- The number of equivalents of reactants and products are equal in a chemical reaction.

Quick Tips:
- Always ensure that the units are consistent before performing calculations.
- Pay attention to the valency of the metal when calculating the atomic weight.

113 Medium

113 of 272

Analyzing +4 | -1

Assertion: If

200 \mathrm{~mL}

of

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

is added to

200 \mathrm{~mL}

of

0.1 \mathrm{~N} \, \mathrm{H}_2 \mathrm{SO}_4

solution, then the resulting solution is acidic.

Reason: If milliequivalents of acid is greater than milliequivalents of base, then upon mixing the solution is acidic.

View answer and explanation

Correct option: D

Assertion is false, Reason is true

Option D is correct because the milliequivalents of NaOH and H2SO4 are equal, resulting in a neutral solution, making the assertion false, but the reason is true.

Solution approach

Correct Answer + Core Reason: Assertion is false, Reason is true. The assertion is false because the resulting solution is neutral, not acidic, as the milliequivalents of acid and base are equal. The reason is true because the acidity of the solution depends on the relative milliequivalents of acid and base.

Step-by-Step Solution:

Step 1: Calculate the milliequivalents of NaOH: Milliequivalents = Normality x Volume (in mL) = 0.1 N x 200 mL = 20 meq

Step 2: Calculate the milliequivalents of H2SO4: Milliequivalents = Normality x Volume (in mL) = 0.1 N x 200 mL = 20 meq

Step 3: Compare the milliequivalents: Since the milliequivalents of NaOH and H2SO4 are equal (20 meq = 20 meq), the resulting solution will be neutral.

Step 4: Analyze the Assertion: The assertion states that the resulting solution is acidic. This is incorrect because the solution is neutral.

Step 5: Analyze the Reason: The reason states that if milliequivalents of acid are greater than milliequivalents of base, then upon mixing, the solution is acidic. This is correct.

Step 6: Conclusion: The assertion is false, but the reason is true.

Why other options are wrong:
- Option A: Incorrect because the assertion is false.
- Option B: Incorrect because the assertion is false.
- Option C: Incorrect because the assertion is false.

Key Concepts:
- Normality
- Milliequivalents
- Acid-base neutralization
- Acidic, basic, and neutral solutions

Memory Aids:
- Milliequivalents = Normality x Volume (in mL)
- Acid + Base -> Salt + Water

Quick Tips:
- Calculate the milliequivalents of acid and base.
- Compare the milliequivalents to determine the nature of the resulting solution.
- If meq(acid) > meq(base), the solution is acidic.
- If meq(base) > meq(acid), the solution is basic.
- If meq(acid) = meq(base), the solution is neutral.

114 Medium

114 of 272

Application +4 | -1

If a piece of copper weighs

0.635 \mathrm{~g}

, how many atoms does it contain?

View answer and explanation

Correct option: A

6.023 \times 10^{21}

Option A (

6.023 \times 10^{21}

) is correct because it is calculated by dividing the given mass by the molar mass and multiplying by Avogadro's number.

Solution approach

Correct Answer + Core Reason: $6.023 \times 10^{21}$ . The number of atoms is calculated by dividing the given mass by the molar mass and multiplying by Avogadro's number.

Step-by-Step Solution:

Step 1: Identify the given information:\nMass of copper = 0.635 g\nMolar mass of copper (Cu) = 63.5 g/mol\nAvogadro's number ( $N_A$ ) = $6.023 \times 10^{23}$ atoms/mol

Step 2: Calculate the number of moles of copper:\nMoles of Cu = $\frac{\text{Mass of Cu}}{\text{Molar mass of Cu}} = \frac{0.635 \mathrm{~g}}{63.5 \mathrm{~g/mol}} = 0.01 \mathrm{~mol}$

Step 3: Calculate the number of atoms of copper:\nNumber of atoms = Moles of Cu $\times$ Avogadro's number\nNumber of atoms = $0.01 \mathrm{~mol} \times 6.023 \times 10^{23} \mathrm{~atoms/mol} = 6.023 \times 10^{21} \mathrm{~atoms}$

Why other options are wrong:
- **Option B ( $6.023 \times 10^{22}$ ): This is incorrect because it is off by a factor of 10. It might be a calculation error.
- Option C ( $6.033 \times 10^{21}$ ): This is incorrect because it uses a slightly different value for Avogadro's number or has a rounding error.
- Option D ( $6.023 \times 10^{25}$ ): This is incorrect because it multiplies the moles by 100 instead of dividing by 100.

Key Concepts:
- Molar Mass
- Avogadro's Number
- Mole Concept

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.

115 Medium

115 of 272

Application +4 | -1

An element

(\mathrm{X})

reacts with hydrogen leading to formation of a class of compounds that is analogous to hydrocarbons.

5 \mathrm{~g}

of

\mathrm{X}

forms

5.628 \mathrm{~g}

of a mixture of two compounds of

\mathrm{X}\left(\mathrm{XH}_4\right.

and

\left.\mathrm{X}_2 \mathrm{H}_6\right)

in the molar ratio of

2: 1.

Determine the molar mass of

\mathrm{X}.

View answer and explanation

Correct option: A

28

Option A (28) is correct because applying the Principle of Atomic Conservation (POAC) on element X allows us to determine its molar mass.

Solution approach

Correct Answer + Core Reason: 28. The molar mass of X is determined by applying POAC on X and solving the resulting equations.

Step-by-Step Solution:

Step 1: The reaction is $\mathrm{X} \rightarrow \mathrm{X}_2 \mathrm{H}_6 + \mathrm{XH}_4$ .

Step 2: Apply POAC on X: $1 \times n_{\mathrm{X}} = 2 \times n_{\mathrm{X}_2 \mathrm{H}_6} + 1 \times n_{\mathrm{XH}_4}$ .

Step 3: Let the molar mass of X be M. Then, $n_{\mathrm{X}} = \frac{5}{M}$ . Also, $n_{\mathrm{XH}_4} = 2 \times n_{\mathrm{X}_2 \mathrm{H}_6}$ .

Step 4: Substitute into the POAC equation: $\frac{5}{M} = 2 \times n_{\mathrm{X}_2 \mathrm{H}_6} + 2 \times n_{\mathrm{X}_2 \mathrm{H}_6} = 4 \times n_{\mathrm{X}_2 \mathrm{H}_6}$ . Therefore, $n_{\mathrm{X}_2 \mathrm{H}_6} = \frac{5}{3M}$ .

Step 5: The total mass of the mixture is 5.628 g. So, $n_{\mathrm{X}_2 \mathrm{H}_6} \times (2M + 6) + n_{\mathrm{XH}_4} \times (M + 4) = 5.628$ .

Step 6: Substitute $n_{\mathrm{XH}_4} = 2 \times n_{\mathrm{X}_2 \mathrm{H}_6}$ into the equation: $n_{\mathrm{X}_2 \mathrm{H}_6} \times (2M + 6) + 2n_{\mathrm{X}_2 \mathrm{H}_6} \times (M + 4) = 5.628$ .

Step 7: Substitute $n_{\mathrm{X}_2 \mathrm{H}_6} = \frac{5}{3M}$ into the equation: $\frac{5}{3M} \times (2M + 6) + \frac{10}{3M} \times (M + 4) = 5.628$ .

Step 8: Simplify the equation: $\frac{10M + 30 + 10M + 40}{3M} = 5.628 \Rightarrow \frac{20M + 70}{3M} = 5.628 \Rightarrow 20M + 70 = 16.884M \Rightarrow 3.116M = 70 \Rightarrow M = \frac{70}{3.116} \approx 28$ .

Step 9: Final Answer: The molar mass of X is approximately 28 g/mol.

Why other options are wrong:
- Option B (58): This value is incorrect as it does not satisfy the mass balance and POAC equations.
- Option C (72): This value is incorrect as it does not satisfy the mass balance and POAC equations.
- Option D (83): This value is incorrect as it does not satisfy the mass balance and POAC equations.

Key Concepts:
- Principle of Atomic Conservation (POAC)
- Mole concept
- Molar mass

Memory Aids:
- POAC: Atoms are neither created nor destroyed in a chemical reaction.

Quick Tips:
- Always start with a balanced chemical equation or a clear understanding of the stoichiometry.
- Apply POAC to each element involved in the reaction.

116 Medium

116 of 272

Application +4 | -1

Rest mass of 1 mole neutrons

\left(m_{\mathrm{n}}=1.675 \times 10^{-27} \mathrm{~kg}\right)

is:

View answer and explanation

Correct option: D

1.008 \times 10^{-3} \mathrm{~kg}

Option D (

1.008 \times 10^{-3} \mathrm{~kg}

) is correct because it is the product of the mass of one neutron and Avogadro's number.

Solution approach

Correct Answer + Core Reason: $1.008 \times 10^{-3} \mathrm{~kg}$ . The rest mass of 1 mole of neutrons is calculated by multiplying the mass of one neutron by Avogadro's number.

Step-by-Step Solution:
Step 1: Recall Avogadro's number: $N_A = 6.023 \times 10^{23} \mathrm{~mol^{-1}}$ .
Step 2: Multiply the mass of one neutron by Avogadro's number to find the mass of one mole of neutrons: $m_{1 \, mole} = m_n \times N_A$ .
Step 3: Substitute the given values: $m_{1 \, mole} = (1.675 \times 10^{-27} \mathrm{~kg}) \times (6.023 \times 10^{23} \mathrm{~mol^{-1}})$ .
Step 4: Calculate the mass: $m_{1 \, mole} = 1.675 \times 6.023 \times 10^{-4} \mathrm{~kg} = 10.088525 \times 10^{-4} \mathrm{~kg} \approx 1.008 \times 10^{-3} \mathrm{~kg}$ .

Why other options are wrong:
- **Option A ( $1.8 \times 10^{-3} \mathrm{~kg}$ ): This is incorrect. It might be obtained by incorrectly multiplying the mass of one neutron by a wrong factor or making a mistake in the calculation.
- Option B ( $1.008 \times 10^{-4} \mathrm{~kg}$ ): This is incorrect. It could result from an error in the exponent or a mistake in the calculation.
- Option C ( $1.08 \times 10^{-3} \mathrm{~kg}$ ): This is incorrect. It may arise from using an incorrect value for Avogadro's number or making a mistake in the calculation.

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

Memory Aids:**
- Remember Avogadro's number: $N_A = 6.023 \times 10^{23} \mathrm{~mol^{-1}}$ .

Quick Tips:
- Always use the correct units for all constants and variables.
- Remember to multiply the mass of one particle by Avogadro's number to find the mass of one mole of particles.

117 Medium

117 of 272

Application +4 | -1

Number of moles of hydrogen atoms present in

60 \mathrm{~amu}

of acetic acid:

View answer and explanation

Correct option: A

4 / \mathrm{N}_{\mathrm{A}}

Option A (

4 / N_A

) is correct because one molecule of acetic acid contains 4 hydrogen atoms, and we are dealing with one molecule (60 amu).

Solution approach

Correct Answer + Core Reason: $4 / N_A$ . Since 60 amu corresponds to one molecule of acetic acid, and each molecule has 4 hydrogen atoms, the number of moles of hydrogen atoms is 4/NA.

Step-by-Step Solution:

Step 1: Identify the molecular formula of acetic acid: $CH_3COOH$ .

Step 2: Determine the number of hydrogen atoms in one molecule of acetic acid: There are 3 hydrogen atoms in the $CH_3$ group and 1 hydrogen atom in the $COOH$ group, for a total of 4 hydrogen atoms.

Step 3: Relate amu to molecules: 60 amu corresponds to one molecule of acetic acid.

Step 4: Calculate the number of moles of hydrogen atoms: Since there are 4 hydrogen atoms in one molecule, there are 4 atoms of H per molecule of acetic acid. To convert to moles, divide by Avogadro's number ( $N_A$ ). Therefore, the number of moles of hydrogen atoms is $\frac{{4}}{{N_A}}$ .

Why other options are wrong:
- Option B: 4 This is incorrect because it represents the number of hydrogen atoms, not the number of moles of hydrogen atoms.
- Option C: 1 This is incorrect because it assumes only one hydrogen atom per molecule of acetic acid.
- Option D: $1 / N_A$ This is incorrect because it assumes only one hydrogen atom per molecule of acetic acid and expresses it in moles.

Key Concepts:
- Molecular formula
- Atomic mass unit (amu)
- Avogadro's number ( $N_A$ )
- Mole concept

Memory Aids:
- Remember that 1 amu is the mass of one atom or molecule.
- Avogadro's number ( $N_A$ ) relates the number of particles to moles.

Quick Tips:
- Always identify the correct molecular formula to count the number of atoms of each element.
- Remember to divide by Avogadro's number to convert from number of atoms/molecules to moles.

118 Medium

118 of 272

Application +4 | -1

1.00 \times 10^{-3}

moles of

{Ag}^{+}

and

1.00 \times 10^{-3}

moles of

{CrO}_4^{2-}

react together to form solid

{Ag}_2 {CrO}_4

. Calculate the amount of

{Ag}_2 {CrO}_4

formed

\left({Ag}_2 {CrO}_4=331.73 \mathrm{~g} \mathrm{~mol}^{-1}\right)

.

View answer and explanation

Correct option: B

0.166 \mathrm{~g}

Option B (

0.166 \mathrm{~g}

) is correct because it is the mass of Ag₂CrO₄ formed, calculated using the limiting reagent concept.

Solution approach

Correct Answer + Core Reason: $0.166 \mathrm{~g}$ . The amount of $Ag_2CrO_4$ formed is determined by the limiting reagent.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2Ag^+ + CrO_4^{2-} \rightarrow Ag_2CrO_4$ .

Step 2: Identify the given information:
- Moles of $Ag^+ = 1.00 \times 10^{-3} \mathrm{~mol}$
- Moles of $CrO_4^{2-} = 1.00 \times 10^{-3} \mathrm{~mol}$
- Molar mass of $Ag_2CrO_4 = 331.73 \mathrm{~g/mol}$

Step 3: Determine the limiting reagent:
- From the balanced equation, 2 moles of $Ag^+$ react with 1 mole of $CrO_4^{2-}$ .
- Moles of $CrO_4^{2-}$ required to react with $1.00 \times 10^{-3} \mathrm{~mol}$ of $Ag^+ = \frac{1}{2} \times 1.00 \times 10^{-3} = 0.50 \times 10^{-3} \mathrm{~mol}$ .
- Since we have $1.00 \times 10^{-3} \mathrm{~mol}$ of $CrO_4^{2-}$ , which is more than required, $Ag^+$ is the limiting reagent.

Step 4: Calculate the moles of $Ag_2CrO_4$ formed:
- Moles of $Ag_2CrO_4$ formed = $\frac{1}{2} \times$ moles of $Ag^+ = \frac{1}{2} \times 1.00 \times 10^{-3} = 0.50 \times 10^{-3} \mathrm{~mol}$ .

Step 5: Calculate the mass of $Ag_2CrO_4$ formed:
- Mass of $Ag_2CrO_4 = \text{moles} \times \text{molar mass} = 0.50 \times 10^{-3} \mathrm{~mol} \times 331.73 \mathrm{~g/mol} = 0.165865 \mathrm{~g} \approx 0.166 \mathrm{~g}$ .

Why other options are wrong:
- **Option A ( $0.268 \mathrm{~g}$ ): Incorrect due to a calculation error or not using the limiting reagent concept correctly.
- Option C ( $0.212 \mathrm{~g}$ ): Incorrect due to a calculation error or not using the limiting reagent concept correctly.
- Option D ( $1.66 \mathrm{~g}$ ): Incorrect due to a calculation error or not using the limiting reagent concept correctly.

Key Concepts:
- Limiting reagent
- Stoichiometry
- Mole concept

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

Quick Tips:**
- Write the balanced chemical equation.
- Calculate the moles of reactants.
- Determine the limiting reagent.
- Calculate the moles of product formed using the limiting reagent.

119 Medium

119 of 272

Application +4 | -1

Number of molecules in 1 litre of oxygen at NTP is:

View answer and explanation

Correct option: B

\frac{6.02 \times 10^{23}}{22.4}

Option B is correct because at NTP, 1 mole of any gas occupies 22.4 L, so the number of molecules in 1 L is Avogadro's number divided by 22.4.

Solution approach

Correct Answer + Core Reason: $\frac{{6.02 \times 10^{23}}}{{22.4}}$ . At NTP, 1 mole of oxygen occupies a volume of 22.4 L.

Step-by-Step Solution:

Step 1: State the given information: Volume = 1 L, Condition = NTP

Step 2: Recall that at NTP, 1 mole of any gas occupies 22.4 L.

Step 3: Calculate the number of moles of oxygen in 1 L: $\frac{{1 \text{{ L}}}}{{22.4 \text{{ L/mol}}}} = \frac{{1}}{{22.4}} \text{{ mol}}$

Step 4: Calculate the number of molecules using Avogadro's number ( $N_A = 6.02 \times 10^{23}$ molecules/mol): Number of molecules = $\frac{{1}}{{22.4}} \text{{ mol}} \times 6.02 \times 10^{23} \text{{ molecules/mol}}$

Step 5: Simplify the expression: Number of molecules = $\frac{{6.02 \times 10^{23}}}{{22.4}}$

Final Answer: $\frac{{6.02 \times 10^{23}}}{{22.4}}$

Why other options are wrong:
- Option A: This option divides Avogadro's number by the molar mass of oxygen (32 g/mol), which is incorrect for finding the number of molecules in 1 L at NTP.
- Option C: This option multiplies the molar mass of oxygen by the molar volume at NTP, which is not relevant to the calculation.
- Option D: This option divides the molar mass of oxygen by the molar volume at NTP, which is also incorrect.

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

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

Quick Tips:
- Always remember the standard molar volume at NTP (22.4 L/mol).

120 Medium

120 of 272

Analyzing +4 | -1

100 \mathrm{~mL}

of

10 \% \mathrm{~NaOH}(w / \mathrm{V})

is added to

100 \mathrm{~mL}

of

10 \% \mathrm{HCl}(w / \mathrm{V})

. The resultant solution becomes:

View answer and explanation

Correct option: C

acidic

Option C (acidic) is correct. The number of moles of HCl is greater than the number of moles of NaOH, resulting in an acidic solution.

Solution approach

Correct Answer + Core Reason: acidic. Moles of HCl are greater than moles of NaOH.

Step-by-Step Solution:

Step 1: Calculate the mass of NaOH in 100 mL of 10% (w/V) solution:
Mass of NaOH = 10% of 100 mL = 10 g

Step 2: Calculate the number of moles of NaOH:
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Moles of NaOH = $\frac{{10 \text{{ g}}}}{{40 \text{{ g/mol}}}} = 0.25 \text{{ mol}}$

Step 3: Calculate the mass of HCl in 100 mL of 10% (w/V) solution:
Mass of HCl = 10% of 100 mL = 10 g

Step 4: Calculate the number of moles of HCl:
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol
Moles of HCl = $\frac{{10 \text{{ g}}}}{{36.5 \text{{ g/mol}}}} \approx 0.274 \text{{ mol}}$

Step 5: Compare the number of moles of NaOH and HCl:
Moles of HCl (0.274 mol) > Moles of NaOH (0.25 mol)

Step 6: Determine the nature of the resultant solution: Since there are more moles of HCl than NaOH, the resultant solution will be acidic.

Why other options are wrong:
- Option A: Incorrect because the solution is acidic, not alkaline.
- Option B: Incorrect because the solution is acidic, not strongly alkaline.
- Option D: Incorrect because the solution is acidic, not neutral.

Key Concepts:
- Molarity
- Acid-Base Neutralization

Memory Aids:
- Acid + Base -> Salt + Water

Quick Tips:
- Always convert concentrations to moles before comparing acidity and basicity.
- Remember the molar masses of common acids and bases.

Your Progress

Some basic concepts of chemistry practice · Medium

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice