A solution is prepared by mixing 8.5 g of CH$$_2$$Cl$$_2$$ and 11.95 g of CHCl$$_3$$. If vapour pressure of CH$$_2$$Cl$$_2$$ and CHCl$$_3$$ at 298K are 415 and 200 mm Hg respectively, the mole fraction of CHCl$$_3$$ in vapour form is: (Molar mass of Cl = 35.5 g mol$$^{-1}$$)

We begin by writing down all the data given in the statement of the problem. The mass of dichloromethane (written as $$CH_2Cl_2$$) is 8.5 g and the mass of chloroform (written as $$CHCl_3$$) is 11.95 g. Their vapour pressures at $$298\text{ K}$$ in the pure state are respectively $$415\;\text{mm Hg}$$ for $$CH_2Cl_2$$ and $$200\;\text{mm Hg}$$ for $$CHCl_3$$.

In order to apply Raoult’s law, we must first convert the given masses into number of moles. So, we calculate the molar masses.

For $$CH_2Cl_2$$:

We have one carbon atom ($$12\;g\;mol^{-1}$$), two hydrogen atoms ($$1\times2=2\;g\;mol^{-1}$$), and two chlorine atoms ($$35.5\times2=71\;g\;mol^{-1}$$). Therefore,

$$M_{CH_2Cl_2}=12+2+71=85\;g\;mol^{-1}.$$

For $$CHCl_3$$:

We have one carbon atom ($$12\;g\;mol^{-1}$$), one hydrogen atom ($$1\;g\;mol^{-1}$$) and three chlorine atoms ($$35.5\times3=106.5\;g\;mol^{-1}$$). Therefore,

$$M_{CHCl_3}=12+1+106.5=119.5\;g\;mol^{-1}.$$

Now we obtain the number of moles present.

For $$CH_2Cl_2$$:

$$n_{CH_2Cl_2}=\dfrac{8.5\;g}{85\;g\;mol^{-1}}=0.1\;mol.$$

For $$CHCl_3$$:

$$n_{CHCl_3}=\dfrac{11.95\;g}{119.5\;g\;mol^{-1}}=0.1\;mol.$$

The total number of moles in the liquid mixture is therefore

$$n_{\text{total}} = 0.1+0.1 = 0.2\;mol.$$

Hence the mole fractions in the liquid phase are easily found.

For $$CH_2Cl_2$$:

$$X_{CH_2Cl_2}=\dfrac{0.1}{0.2}=0.5.$$

For $$CHCl_3$$:

$$X_{CHCl_3}=\dfrac{0.1}{0.2}=0.5.$$

Raoult’s law states that the partial vapour pressure of each component in an ideal solution is equal to the product of its mole fraction in the liquid phase and the vapour pressure of the pure component. Symbolically, for a component $$i$$ we write

$$P_i = X_i \, P_i^{\,0},$$

where $$P_i^{\,0}$$ is the vapour pressure of the pure liquid.

Applying this to the two liquids, we get

For $$CH_2Cl_2$$:

$$P_{CH_2Cl_2}=X_{CH_2Cl_2}\,P^{0}_{CH_2Cl_2}=0.5\times415\;\text{mm Hg}=207.5\;\text{mm Hg}.$$

For $$CHCl_3$$:

$$P_{CHCl_3}=X_{CHCl_3}\,P^{0}_{CHCl_3}=0.5\times200\;\text{mm Hg}=100\;\text{mm Hg}.$$

The total vapour pressure of the solution is the sum of these two partial pressures, so

$$P_{\text{total}} = P_{CH_2Cl_2}+P_{CHCl_3}=207.5+100 = 307.5\;\text{mm Hg}.$$

To find the composition of the vapour phase, we now use Dalton’s law. The mole fraction of any component in the vapour phase equals its partial pressure divided by the total pressure. Thus, the mole fraction of $$CHCl_3$$ in the vapour is

$$y_{CHCl_3}= \dfrac{P_{CHCl_3}}{P_{\text{total}}}= \dfrac{100}{307.5}.$$

Carrying out the division, we obtain

$$y_{CHCl_3}=0.3252\;(\text{approximately}).$$

Rounding to the number of significant figures implied in the given options, we can write

$$y_{CHCl_3}\approx0.325.$$

Hence, the correct answer is Option C.