Question

A solution is prepared by mixing 8.5g of $C{H_2}C{l_2}$ and 11.95g of $CHC{l_3}$.If vapour pressure of $C{H_2}C{l_2}$ and $CHC{l_3}$ at 4298K are 415 and 200mmHg respectively, the mole fraction of $CHC{l_3}$ in vapour form is : (molar mass of Cl = $35.5gmo{l^{ - 1}}$).
A. 0.162
B. 0.675
C. 0.325
D. 0.456

Answer 1

In order to solve this question we have to find mole fraction and then mole fraction at vapour phase. First we will find the mole fraction by dividing the given weight by atomic mass then we will calculate the partial pressure constituted by each of them. At last we can easily calculate the mole fraction at vapour phase by knowing mole fraction and partial pressure.

Complete step by step solution:
Molar mass of $C{H_2}C{l_2}$, ${M_1}$= $85g{(mol)^{ - 1}}$
Moles of C{H_2}C{l_2}$$$({n_1})$=$\dfrac{{{W_1}}}{{{M_1}}} = \dfrac{{8.5g}}{{85g{{(mol)}^{ - 1}}}} = 0.1mol$ Molar mass of $CHC{l_3}$,${M_2} = 119.5g{(mol)^{ - 1}}$ Moles of $CHC{l_3}$ $({x_1})$=$\dfrac{{W{}_2}}{{{M_2}}} = \dfrac{{11.95g}}{{119.5g{{(mol)}^{ - 1}}}} = 0.1mol$ Mole fraction of $CHC{l_3}$ $({x_2})$=$\dfrac{{{n_2}}}{{{n_1} + {n_2}}} = \dfrac{{0.1mol}}{{0.2mol}} = 0.5$ Partial vapour pressure of $CHC{l_3}$ $({p_2}) = {x_2}p_2^0 = 0.5 \times 200mmHg = 100mmHg$ Partial vapour pressure of C{H_2}C{l_2}$$$({p_1}) = (1 - {x_2})p_1^0 = 0.5 \times 415mmHg = 207.5mmHg$Mole fraction of$CHC{l_3}$in vapour phase is${y_2} = \dfrac{{{p_2}}}{{{p_1} + {p_2}}} = \dfrac{{100mmHg}}{{307.5mmHg}} = 0.325$.

Hence, the correct option is (C).

Note: The number of molecules of a particular component in a mixture divided by the total number of moles in the given mixture, is a mole fraction. Whereas, the molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles.