Question

Two volatile liquids A and B are mixed together in a ratio of $3:2$ by mole to form an ideal solution.
Calculate the total pressure of the solution
Given:
$\text{Vapour pressure of pure A,}\left(p_A^{\circ }\right)=200mmHg$
$\text{Vapour pressure of pure B,}\left(p_B^{\circ }\right)=50mmHg$

• A. $100mmHg$
• B. $110mmHg$
• C. $130mmHg$
• D. $140mmHg$

Answer 1

The correct option is D $140mmHg$

$\text{Mole fraction of A =}\frac{\text{No. of moles of A}}{\text{Total no. of moles of A and B}}$

$x_A=\frac{n_A}{n_A+n_B}=\frac{3}{2+3}=\frac{3}{5}$

$x_B=1-\frac{3}{5}=\frac{2}{5}$
Partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.
$\text{By Raoult's law,}$
$p_A=x_A\times p_A^0$
$p_B=x_B\times p_B^0$
$P_{total}=x_A\times p_A^0+x_B\times p_B^0=200\times \frac{3}{5}+50\times \frac{2}{5}=120+20=140mmHg$