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

Concept: Mole fraction from mass percentage

When mass percentage is given, convert masses to moles using molar masses, then apply the mole fraction definition.

Solution:

Assume $100 \text{ g}$ of solution. Then benzene = $30 \text{ g}$ and carbon tetrachloride = $70 \text{ g}$.

Molar mass of benzene ($\text{C}_6\text{H}_6$) = $78 \text{ g mol}^{-1}$

Molar mass of carbon tetrachloride ($\text{CCl}_4$) = $154 \text{ g mol}^{-1}$

Moles of benzene: $n_{\text{benzene}} = \frac{30}{78} = 0.385 \text{ mol}$

Moles of $\text{CCl}_4$: $n_{\text{CCl}_4} = \frac{70}{154} = 0.455 \text{ mol}$

Mole fraction of benzene:

Basis: 100 g of solution. 30% by mass benzene $\Rightarrow$ 30 g benzene and 70 g carbon tetrachloride.

Moles. Molar mass of benzene $\mathrm{C_6H_6}=78\ \text{g mol}^{-1}$; of $\mathrm{CCl_4}=154\ \text{g mol}^{-1}$:

Mole fraction of benzene.

Mole Fraction of Benzene in a Solution with Carbon Tetrachloride

1. Concept First — Mass Percentage to Mole Fraction

This problem tests your ability to convert mass percentage into mole fraction — a fundamental skill in solution chemistry. The key idea is:

• Mass percentage tells us the mass of each component in 100 g of solution.
• Mole fraction tells us the ratio of moles of one component to total moles.

The intuition: Even though we're given mass, chemistry happens in moles (particles). So we must convert mass → moles using molar masses, then find the fraction. (Mole fraction is also the quantity later chapters and laws — such as Raoult's law — work with, which is why this conversion skill matters.)

2. Step-by-Step Solution

Step 1: Interpret the given data

We have a solution containing 30% by mass of benzene in carbon tetrachloride ($\text{CCl}_4$).

This means:

• In 100 g of solution:
  • Mass of benzene = 30 g
  • Mass of carbon tetrachloride = 100 g − 30 g = 70 g

Step 2: Find molar masses

We need the molar masses of both substances:

• Benzene ($\text{C}_6\text{H}_6$):

  • Carbon: $6 \times 12 = 72$
  • Hydrogen: $6 \times 1 = 6$
  • Molar mass = 78 g/mol

• Carbon tetrachloride ($\text{CCl}_4$):

  • Carbon: $1 \times 12 = 12$
  • Chlorine: $4 \times 35.5 = 142$
  • Molar mass = 154 g/mol

Step 3: Calculate moles of each component

Using the formula: $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$

• Moles of benzene:

• Moles of carbon tetrachloride:

Step 4: Calculate total moles

Step 5: Calculate mole fraction of benzene

Mole fraction is defined as:

3. Final Answer

(Rounded to three significant figures. NCERT's answer key prints 0.459 — a last-digit difference from rounding at intermediate steps; the fully-unrounded computation gives 0.458.)

4. Why It Works & Exam Tip

Why this approach works:

• Mass percentage gives a convenient 100 g sample to work with.
• Converting to moles is essential because mole fraction is a mole-based quantity.
• The calculation is simply: moles of benzene ÷ total moles.

Common pitfall to avoid:

✗ Do not directly use mass ratio as mole fraction.

For example, don't write $\frac{30}{100} = 0.3$ as the mole fraction — that's the mass fraction, not mole fraction.

✓ Always convert to moles first — different substances have different molar masses, so equal masses do not mean equal moles.

Quick check:

Since benzene has a lower molar mass (78) than CCl₄ (154), 30 g of benzene gives more moles than you might expect from mass alone. That's why the mole fraction (0.458) is higher than the mass fraction (0.30).

Common Mistakes in Converting Mass Percentage to Mole Fraction

1. Using Mass Fraction Directly as Mole Fraction

The Mistake: Writing mole fraction of benzene = 30/100 = 0.30.

Why it's wrong: 30% by mass is a mass ratio; mole fraction requires converting to moles first, since benzene and CCl4 have different molar masses.

How to avoid: Always convert mass to moles before computing any fraction.

2. Wrong Molar Mass of CCl4

The Mistake: Using CCl4's molar mass as 12+35.5=47.5 (forgetting there are 4 chlorine atoms).

How to avoid: $M_{CCl_4}=12+4(35.5)=154\,\text{g/mol}$.

3. Forgetting the 100 g Basis

The Mistake: Not realizing '30% by mass' means 30 g benzene per 100 g of solution, so the solvent mass is 70 g, not 100 g.

How to avoid: Always write: mass of solute + mass of solvent = 100 g when given a mass percentage with no absolute mass stated.

Correct Solution (for reference)

Basis: 100 g solution -> 30 g benzene, 70 g CCl4.

Final Answer: Mole fraction of benzene ≈ 0.458.

(NCERT's answer key prints 0.459 for benzene and 0.541 for CCl₄ — a last-digit difference from rounding at intermediate steps; the fully-unrounded computation gives 0.458 and 0.542.)