Question

A solution is prepared by mixing 8.5 g of ch2cl2 and 11.95 g of chcl3 . If vapour pressure of ch2cl2 and chcl3 at 298 k are 415 and 200 mm hg respectively, the mole fraction of chcl3 in vapour form is : (molar mass of cl = 35.5 g mol1)

Answer 1

the mole fraction of CHCl3 in vapour form is 

0.162

Answer 2

Answer: Thus the mole fraction of $CHCl_3$ is 0.32.

Explanation: According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_1=x_1p_1^0$ and $p_2=x_2P_2^0$

where, x = mole fraction

$p^0$ = pressure in the pure state

Now we have to calculate the moles of $CH_2Cl_2$ and $CHCl_3$

$\text{Moles of }CH_2Cl_2=\frac{\text{Mass of }CH_2Cl_2}{\text{Molar mass of }CH_2Cl_2}=\frac{8.5g}{85g/mole}=0.1moles$

$\text{Moles of }CHCl_3=\frac{\text{Mass of }CHCl_3}{\text{Molar mass of }CHCl_3}=\frac{11.95g}{119.4g/mole}=0.1moles$

Now wee have to calculate the mole fraction

$x_{CH_2Cl_2}=\frac{n_{CH_2Cl_2}}{n_{CH_2Cl_2}+n_{CHCl_3}}$

$x_{CH_2Cl_2}=\frac{0.1}{0.1+0.1}=0.5$

$x_{CHCl_3}=\frac{0.1}{0.1+0.1}=0.5$

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$

$p_{total}=x_{CH_2Cl_2}p_{CH_2Cl_2}^0+x_{CHCl_3}p_{CHCl_3}^0$

$p_{CH_2Cl_2}^0=415mmHg$

$p_{CHCl_3}^0=200mmHg$

$p_{total}=0.5\times 415+0.5\times 200=307.5mmHg$

The mole fraction of $CHCl_3$ in vapor phase is given by:

$y_{CHCl_3}=\frac{p_{CHCl_3}}{P_{total}}$

$y_{CHCl_3}=\frac{100}{307.5}=0.32$

Thus the mole fraction of $CHCl_3$ is 0.32.