Ncert Exercises & Solutions

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

Let the total mass of the solution be 100 g and the mass of benzene be 30 g.

Mass of carbon tetrachloride = (100 − 30)g = 70 g

Molar mass of benzene (C₆H₆) = (6 × 12 + 6 × 1) g mol⁻¹ = 78 g mol⁻¹

$∴ Number of moles of C_{6} H_{6} = \frac{30}{78} mol = 0.3846 mol$

Molar mass of carbon tetrachloride (CCl₄) = 1 × 12 + 4 × 355 = 154 g mol⁻¹

$∴ Number of moles of {CCl}_{4} = \frac{70}{154} mol = 0.4545 mol$

Thus, the mole fraction of C6H6 is given as:

$\frac{Number of moles of C_{6} H_{6}}{Number of moles of C_{6} H_{6} + Number of moles of {CCl}_{4}}$
$= \frac{0.3846}{0.3846 + 0.4545}$
$= 0.458$