Question

Two liquids A and B having molecular masses 50 and 100 respectively are mixed together in
1:1 mass ratio P' = 600 mm of Hg, PO = 300 mm of Hg. Calculate mole fraction of B in
gaseous state.
(A) 0.8
(C) 0.67
B) 0.2
(D) 0.33​

Answer 1

The liquids are mixed in 1 : 1 mass ratio. Let mass of each liquid mixed be $\tt{m}$ grams.

• Molecular mass of liquid A, $\tt{M_A=50\ g\,mol^{-1}}$

• Molecular mass of liquid B, $\tt{M_B=100\ g\,mol^{-1}}$

• No. of moles of liquid A, $\tt{n_A=\dfrac{m}{50}\ mol}$

• No. of moles of liquid B, $\tt{n_B=\dfrac{m}{100}\ mol}$

Then mole fraction of liquid B in the solution is,

$\tt{\longrightarrow \chi_B=\dfrac{n_B}{n_A+n_B}}$

$\tt{\longrightarrow \chi_B=\dfrac{\left(\dfrac{m}{100}\right)}{\left(\dfrac{m}{50}+\dfrac{m}{100}\right)}}$

$\tt{\longrightarrow \chi_B=\dfrac{\left(\dfrac{m}{100}\right)}{\left(\dfrac{3m}{100}\right)}}$

$\tt{\longrightarrow \chi_B=\dfrac{1}{3}}$

• Vapour pressure of pure liquid A, $\tt{P^0_A=600\ mm\ of\ Hg}$

• Vapour pressure of pure liquid B, $\tt{P^0_B=300\ mm\ of\ Hg}$

Total vapour pressure of the solution,

$\tt{\longrightarrow P_T=P^0_A\chi_A+P^0_B\chi_B}$

$\tt{\longrightarrow P_T=600\cdot\dfrac{2}{3}+300\cdot\dfrac{1}{3}\quad\quad\!\!\left[\because\chi_A=1-\chi_B=\dfrac{2}{3}\right]}$

$\tt{\longrightarrow P_T=500\ mm\ of\ Hg}$

Then the mole fraction of liquid B in gaseous state is given by,

$\tt{\longrightarrow \gamma_B=\dfrac{P^0_B\chi_B}{P_T}}$

$\tt{\longrightarrow \gamma_B=\dfrac{300\cdot\dfrac{1}{3}}{500}}$

$\tt{\longrightarrow\underline{\underline{\gamma_B=0.2}}}$

Hence (B) is the answer.