Question

$100g$ of liquid $A$(molar mass $140g{mol}^{-1})$ was dissolved in $1000g$ of liquid $B$ (molar mass $180g{mol}^{-1})$. The vapour pressure of pure liquid $B$ was found to be $500torr.$ Calculate the vapour of pure liquid $A$ and its vapour pressure in the solution if the total vapour pressure of the solution is $475torr$.

• A. $22torr$
• B. $32torr$
• C. $45torr$
• D. $36torr$

Answer 1

The correct option is B $32torr$

Given:-

$P_T=475$ Torr, $P_B^{\circ }=500$ Torr

${\omega }_A=100g,M_A=140g;{\omega }_B=1000g,M_B=180g$

To find:- $P_A^0$ and $P_A$

$n_A=\frac{100}{140}=\frac{5}{7},n_B=\frac{1000}{180}=\frac{50}{9}$

Now, $x_A=\frac{n_A}{h_A+h_B}=\frac{\frac{5}{7}}{\frac{5}{7}+\frac{50}{9}}=0.114$

$x_B=1-u_A=1-0.114=0.886$

we know, that,

$P_T=P_A^0{x}_A+P_B^0{x}_B$

where $p_A{x}_A=P_A$ and $p_B{x}_B=P_B$

Now, $P^{0}A\times 0.114+500\times 0.886=475$

$\Rightarrow p^{0}A=280\cdot 70$ Torr

$\therefore P_A=p_A{x}_A=280.7\times 0.114$

So, $P_A=31.9998$ Torr

or

$P_A\approx 32$ Torr

option $B$ is correct