The vapour pressure of pure CHCl_3 and CH_2Cl | vedclass

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(C) The molar mass of $CHCl_3$ is $119.5 \,g/mol$ and the molar mass of $CH_2Cl_2$ is $85 \,g/mol$.Number of moles of $CHCl_3$ $(n_1)$ = $\frac{11.9 \

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$94.3$

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(C) The molar mass of $CHCl_3$ is $119.5 \,g/mol$ and the molar mass of $CH_2Cl_2$ is $85 \,g/mol$.Number of moles of $CHCl_3$ $(n_1)$ = $\frac{11.9 \,g}{119.5 \,g/mol} \approx 0.1 \,mol$.Number of moles of $CH_2Cl_2$ $(n_2)$ = $\frac{17 \,g}{85 \,g/mol} = 0.2 \,mol$.Total moles = $0.1 + 0.2 = 0.3 \,mol$.Mole fraction of $CHCl_3$ $(x_1)$ = $\frac{0.1}{0.3} = \frac{1}{3}$.Mole fraction of $CH_2Cl_2$ $(x_2)$ = $\frac{0.2}{0.3} = \frac{2}{3}$.According to Raoult's law,the total vapour pressure $(P_T)$ is:$P_T = P_1^{\circ} x_1 + P_2^{\circ} x_2$$P_T = (200 	imes \frac{1}{3}) + (41.5 	imes \frac{2}{3})$$P_T = 66.67 + 27.67 = 94.34 \,atm \approx 94.3 \,atm$.

The vapour pressure of pure $CHCl_3$ and $CH_2Cl_2$ are $200 \,atm$ and $41.5 \,atm$ respectively. The weights of $CHCl_3$ and $CH_2Cl_2$ are respectively $11.9 \,g$ and $17 \,g$. The vapour pressure of the solution will be (in $,atm$)

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