Answer:
Option A
Explanation:
Given, vapour pressure of pure $CCl_{4}$ at 25° C
$(P^{0}_{CCl_{4}})=$ 115.0 torr
Vapour pressure of pure $SnCl_{4}$ at $25^{0}$C
$(P^{0}_{SnCl_{4}})=$ 238.0 torr
Molar mass of $CCl_{4}$ $ [M_{(CCl_{4})}]=154 g$$mol ^{-1}$
Molar mass of $SnCl_{4}$ in solution $[M_{SnCl_{4}}]$= 170 g $mol^{-1}$
Mass of $CCl_{4}$ in solution $(w_{CCl_{4}})=10g$
mass of $SnCl_{4}$ in solution $(w_{SnCl_{4}})=15 g$
from Dalton's law of partial pressure,
$p_{total}=\chi_{CCl_{4}}.p^{0}_{CCl_{4}}+\chi_{SnCl_{4}}. p^{0}_{SnCl_{4}}$
and $\chi=\frac{n_{simple}}{n_{total}}=\frac{w}{M \times n_{total}}$
Now, $n=\left[\frac{w}{M}\right]_{CCl_{4}}=\frac{10}{154}=0.065$
and $\left[\frac{w}{M}\right]_{SnCl_{4}}=\frac{15}{170}=0.09$
Thus, $\chi_{CCl_{4}}=\frac{n_{CCl_{4}}}{n_{CCl_{4}}+n_{SnCl_{4}}}$
= $\frac{0.065}{0.06+0.09}$
$\chi_{CCl_{4}}=\frac{0.065}{0.15}=0.43$
Therefore, $\chi_{SnCl_{4}}=1-0.43=0.57$
Thus, $p_{total}$= $0.43 \times 115.00+0.57 \times 238$
=49.45+135.66= 185.11≈185.85 torr
Hence, option (a) is the correct answer
