CBSE Class 12 Chemistry Chapter 3: Solutions NCERT Solutions

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NCERT Solutions PDF Class 12 PDF

CBSE Class 12 Chemistry Chapter 3 Solutions offers comprehensive NCERT solutions designed to help students master the concepts of solutions. This guide meticulously explains how to calculate the mass percentage of components, determine the mole fraction of a solute, and find the molarity of various solutions. Each solution breaks down complex calculations into simple, manageable steps, clearly illustrating the formulas used and their practical application. By working through these examples, students can solidify their understanding of different concentration units and calculation methods. This resource is an excellent tool for reinforcing learning, practicing problem-solving skills, and ensuring thorough preparation for board examinations, making the study of solutions more accessible and effective.

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BoardCBSE
ClassClass 12
SubjectChemistry
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 3

Chapter summary

Chapter 3 of the NCERT Class 12 Chemistry syllabus focuses on Solutions. These NCERT Solutions cover in-text questions that help students understand and calculate key concentration terms like mass percentage, mole fraction, and molarity. The exercises involve practical applications of these concepts using given masses and volumes of solutions and solutes.

Learning outcomes

  • Understand the concept of mass percentage and calculate it for binary solutions.
  • Learn to calculate the mole fraction of components in a solution.
  • Apply the definition of molarity to calculate it for given solute mass and solution volume.
  • Solve problems involving dilution of solutions and molarity calculations.
  • Reinforce understanding of molar mass calculations for compounds.

Topics covered

Paper topics

  • Mass Percentage
  • Mole Fraction
  • Molarity
  • Molar Mass Calculation
  • Benzene
  • Carbon Tetrachloride
  • Hydrated Salts
  • Concentration of Solutions

Important topics

  • Mass Percentage Calculation
  • Mole Fraction Calculation
  • Molarity Calculation
  • Understanding Concentration Units

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Questions and Solutions

Question 2.1

Calculate the mass percentage of benzene (C<sub>6</sub>H<sub>6</sub>) and carbon tetrachloride (CCl<sub>4</sub>) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Solution:

To calculate the mass percentage of each component in the solution, we use the formula:

\text{Mass percentage of a component} = \frac{\text{Mass of the component}}{\text{Total mass of the solution}} \times 100\%

Given:

  • Mass of benzene (C<sub>6</sub>H<sub>6</sub>) = 22 g
  • Mass of carbon tetrachloride (CCl<sub>4</sub>) = 122 g

First, calculate the total mass of the solution:

\text{Total mass of solution} = \text{Mass of benzene} + \text{Mass of carbon tetrachloride}

\text{Total mass of solution} = 22 \, \text{g} + 122 \, \text{g} = 144 \, \text{g}

Now, calculate the mass percentage of benzene:

\text{Mass percentage of benzene} = \frac{22 \, \text{g}}{144 \, \text{g}} \times 100\%

\text{Mass percentage of benzene} \approx 15.28\%

Next, calculate the mass percentage of carbon tetrachloride:

\text{Mass percentage of CCl}_4 = \frac{122 \, \text{g}}{144 \, \text{g}} \times 100\%

\text{Mass percentage of CCl}_4 \approx 84.72\%

Alternatively, since there are only two components, the mass percentage of CCl<sub>4</sub> can be calculated as:

\text{Mass percentage of CCl}_4 = (100\% - \text{Mass percentage of benzene})

\text{Mass percentage of CCl}_4 = (100\% - 15.28\%) = 84.72\%

Answer: The mass percentage of benzene is approximately 15.28% and the mass percentage of carbon tetrachloride is approximately 84.72%.

Question 2.2

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Solution:

We are given that the solution contains 30% benzene by mass in carbon tetrachloride. This means that in a 100 g sample of the solution, there are 30 g of benzene and the rest is carbon tetrachloride.

Given:

  • Mass of benzene (C<sub>6</sub>H<sub>6</sub>) = 30 g
  • Total mass of solution = 100 g

Therefore, the mass of carbon tetrachloride (CCl<sub>4</sub>) is:

\text{Mass of CCl}_4 = \text{Total mass of solution} - \text{Mass of benzene}

\text{Mass of CCl}_4 = 100 \, \text{g} - 30 \, \text{g} = 70 \, \text{g}

To calculate the mole fraction, we need to find the number of moles of each component. First, we determine their molar masses:

Molar mass of benzene (C<sub>6</sub>H<sub>6</sub>) = (6 × atomic mass of C) + (6 × atomic mass of H)

= (6 \times 12.011 \, \text{g mol}^{-1}) + (6 \times 1.008 \, \text{g mol}^{-1})

\approx 72.066 \, \text{g mol}^{-1} + 6.048 \, \text{g mol}^{-1} = 78.114 \, \text{g mol}^{-1}

Number of moles of benzene:

n_{\text{C}_6\text{H}_6} = \frac{\text{Mass of benzene}}{\text{Molar mass of benzene}} = \frac{30 \, \text{g}}{78.114 \, \text{g mol}^{-1}} \approx 0.384 \, \text{mol}

Molar mass of carbon tetrachloride (CCl<sub>4</sub>) = (1 × atomic mass of C) + (4 × atomic mass of Cl)

= (1 \times 12.011 \, \text{g mol}^{-1}) + (4 \times 35.453 \, \text{g mol}^{-1})

\approx 12.011 \, \text{g mol}^{-1} + 141.812 \, \text{g mol}^{-1} = 153.823 \, \text{g mol}^{-1}

Number of moles of carbon tetrachloride:

n_{\text{CCl}_4} = \frac{\text{Mass of CCl}_4}{\text{Molar mass of CCl}_4} = \frac{70 \, \text{g}}{153.823 \, \text{g mol}^{-1}} \approx 0.455 \, \text{mol}

The mole fraction of benzene (χ<sub>benzene</sub>) is calculated as:

\chi_{\text{C}_6\text{H}_6} = \frac{\text{Number of moles of benzene}}{\text{Total number of moles in the solution}}

\chi_{\text{C}_6\text{H}_6} = \frac{n_{\text{C}_6\text{H}_6}}{n_{\text{C}_6\text{H}_6} + n_{\text{CCl}_4}}

\chi_{\text{C}_6\text{H}_6} = \frac{0.384 \, \text{mol}}{0.384 \, \text{mol} + 0.455 \, \text{mol}} = \frac{0.384}{0.839}

\chi_{\text{C}_6\text{H}_6} \approx 0.458

Answer: The mole fraction of benzene in the solution is approximately 0.458.

Question 2.3

Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO<sub>3</sub>)<sub>2</sub>. 6H<sub>2</sub>O in 4.3 L of solution (b) 30 mL of 0.5 M H<sub>2</sub>SO<sub>4</sub> diluted to 500 mL.
Solution:

Molarity (M) is defined as the number of moles of solute dissolved per liter of solution. The formula is:

\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}}

(a) Molarity of 30 g of Co(NO<sub>3</sub>)<sub>2</sub>. 6H<sub>2</sub>O in 4.3 L of solution

First, we need to calculate the molar mass of the solute, cobalt(II) nitrate hexahydrate (Co(NO<sub>3</sub>)<sub>2</sub>. 6H<sub>2</sub>O).

Molar mass of Co = 58.93 g mol<sup>-1</sup>

Molar mass of N = 14.01 g mol<sup>-1</sup>

Molar mass of O = 16.00 g mol<sup>-1</sup>

Molar mass of H = 1.01 g mol<sup>-1</sup>

Molar mass of Co(NO<sub>3</sub>)<sub>2</sub>. 6H<sub>2</sub>O = (58.93) + 2 × (14.01 + 3 × 16.00) + 6 × (2 × 1.01 + 16.00)

= 58.93 + 2 \times (14.01 + 48.00) + 6 \times (2.02 + 16.00)

= 58.93 + 2 \times (62.01) + 6 \times (18.02)

= 58.93 + 124.02 + 108.12 = 291.07 \, \text{g mol}^{-1}

Now, calculate the number of moles of the solute:

\text{Moles of Co(NO}_3)_2.6H_2O = \frac{\text{Mass of solute}}{\text{Molar mass of solute}}

= \frac{30 \, \text{g}}{291.07 \, \text{g mol}^{-1}} \approx 0.103 \, \text{mol}

The volume of the solution is given as 4.3 L.

Now, calculate the molarity:

\text{Molarity} = \frac{0.103 \, \text{mol}}{4.3 \, \text{L}} \approx 0.024 \, \text{mol L}^{-1}

Answer (a): The molarity of the solution is approximately 0.024 M.

(b) Molarity of 30 mL of 0.5 M H<sub>2</sub>SO<sub>4</sub> diluted to 500 mL

We can use the dilution formula, M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub>, where:

  • M<sub>1</sub> is the initial molarity
  • V<sub>1</sub> is the initial volume
  • M<sub>2</sub> is the final molarity
  • V<sub>2</sub> is the final volume

Given:

  • Initial molarity (M<sub>1</sub>) = 0.5 M
  • Initial volume (V<sub>1</sub>) = 30 mL
  • Final volume (V<sub>2</sub>) = 500 mL

We need to find the final molarity (M<sub>2</sub>).

Rearranging the formula to solve for M<sub>2</sub>:

M_2 = \frac{M_1 V_1}{V_2}

Substitute the given values:

M_2 = \frac{(0.5 \, \text{M}) \times (30 \, \text{mL})}{500 \, \text{mL}}

M_2 = \frac{15}{500} \, \text{M}

M_2 = 0.03 \, \text{M}

Answer (b): The molarity of the diluted solution is 0.03 M.

Common mistakes

  • Incorrectly calculating the total mass of the solution.
  • Errors in determining the molar masses of compounds.
  • Confusing mass percentage with mole fraction calculations.
  • Mistakes in converting volume units (e.g., mL to L) for molarity calculations.

Revision tips

  • Review the definitions of mass percentage, mole fraction, and molarity before attempting problems.
  • Practice each type of calculation (mass percentage, mole fraction, molarity) separately.
  • Pay close attention to units and ensure consistency throughout calculations.
  • Use the provided molar masses as a reference and practice calculating them for other compounds.

Practice MCQs

Q1. What is the mass percentage of a component if 22 g of it is dissolved in 122 g of another substance?

Q2. If a solution contains 30% benzene by mass, what is the mass of carbon tetrachloride in 100 g of the solution?

Q3. Which of the following is the molar mass of benzene (C6H6)?

Q4. Molarity is defined as:

Frequently asked questions

What are the key concepts covered in these NCERT Solutions for Class 12 Chemistry Chapter 3?

These solutions cover the calculation of mass percentage, mole fraction, and molarity for different solutions, as presented in the in-text questions of Chapter 3.

How do these solutions help in preparing for the CBSE Class 12 Chemistry exam?

They provide clear, step-by-step explanations for solving problems related to solution concentration, helping students understand the methods and practice calculations for exam revision.

What is mass percentage and how is it calculated?

Mass percentage expresses the mass of a component in a solution as a percentage of the total mass of the solution. It is calculated using the formula: (Mass of component / Total mass of solution) × 100%.

How is mole fraction different from mass percentage?

Mole fraction is the ratio of the moles of one component to the total moles of all components in the solution, whereas mass percentage is based on the mass of components relative to the total mass.

What is molarity and what are its units?

Molarity is defined as the number of moles of solute dissolved in one liter of solution. Its units are moles per liter (mol/L).

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