Ncert Exercises & Solutions

2.38 Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Molar mass of benzene $(C_{6} H_{6}) = 6 \times 12 + 6 \times 1 = 78 g {mol}^{- 1}$

Molar mass of toluene $(C_{6} H_{5} {CH}_{3}) = 7 \times 12 + 8 \times 1 = 92 g {mol}^{- 1}$

Now, no. of moles present in 80 g of benzene $= \frac{80}{78} mol = 1.026 mol$

And, no. of moles present in 100 g of toluene $= \frac{100}{92} mol = 1.087 mol$

Mole fraction of benzene, $x_{b} = \frac{1.026}{1.026 + 1.087} = 0.486$

And, mole fraction of toluene, $x_{1} = 1 - 0.486 = 0.514$

It is given that vapour pressure of pure benzene, $p_{b}^{0} = 50.71 mm Hg$

And, vapour pressure of pure toluene, $p_{t}^{0} = 32.06 mm Hg$

Therefore, partial vapour pressure of benzene, $p_{b} = x_{b} \times p_{b} = 0.486 \times 50.71 = 24.645 mm Hg$

And, partial vapour pressure of toluene, $p_{t} = x_{t} \times p_{t} = 0.514 \times 32.06 = 16.479 mm Hg$

Hence, mole fraction of benzene in vapour phase is given by:

$\frac{p_{b}}{p_{b} + p_{t}}$
$= \frac{24.645}{24.645 + 16.479}$
$= \frac{24.645}{41.124}$
$= 0.599$
$= 0.6$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions