Question

100 g of liquid $A$ (molar mass 140 g $mo{l}^{-1}$) was dissolved in 1000 g of liquid $B$ (molar mass 180 g $mo{l}^{-1}$) The vapour pressure of pure liquid $B$ was found to be 500 torr.

Calculate the vapour pressure of pure liquid $A$ and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr:

• A. $3.19,500$
• B. $283.18,31.999$
• C. $500,300$
• D. $190,31.965$

Answer 1

Answer: (B)

$283.18,31.999$

Given: $P_T=475,W_A=100g,M_A=140,P_B^o=500$

$W_B=1000g,M_B=180,P_A^o?,P_A=?$

$P_T=P_A^o{X}_A+P_B^o{X}_B⟶\left(1\right)$ & $X_A+X_B=1⟶\left(2\right)$

$\Rightarrow \frac{X_B}{X_A}=\frac{W_B{M}_A}{W_A{M}_B}=\frac{1000\times 140}{100\times 180}=7.78$

$\therefore X_B=7.78X_A$

Substituting $X_B$ in $\left(2\right)$

$X_A+7.78X_A=1$

$\therefore X_A=0.113$ & hence $X_B=1-0.113=0.886$

Substituting $X_A$ & $X_B$ in equation $\left(1\right)$

$\Rightarrow 475=P_A^o\left(0.113\right)+500\left(0.886\right)$

$\Rightarrow 0.113P_A^o=32\Rightarrow P_A^o=\frac{32}{0.113}=283.18torr$

Now, $P_A=P_A^o(1-X_B)=P_A^o{X}_A$

$=283.18\times 0.113$

$=31.999torr$