Question

If vapour pressure of pure liquid $A$ is $100\text{mm Hg}$ and Vapour pressure of pure liquid $B$ is $150\text{mm Hg}$. $2$ mole of liquid $A$ and $3$ mole of liquid $B$ are mixed to form an ideal solution. The vapour pressure of solution will be:

• A. $135\text{mm}$
• B. $130\text{mm}$
• C. $140\text{mm}$
• D. $145\text{mm}$

Answer 1

The correct option is B $130\text{mm}$

Given,
$\text{Vapour pressure of pure liquid A,}{p}_A^o=100\text{mm Hg}\text{Vapour pressure of pure liquid B},p_B^o=150\text{mm Hg}$
Let, $x_A$ be the mole fraction of liquid $A$ and $x_B$ be the mole fraction of liquid $B$
$x_A=\frac{2}{5}$
$x_B=\frac{3}{5}$
By using Rault's law
$p=p_A^o{x}_A+p_B^o{x}_B$
Substituting the values, we get
$p=100\times \frac{2}{5}+150\times \frac{3}{5}=40+90=130\text{mm}$