Question

3 moles of liquid 'A' and 6 moles of liquid 'B' are mixed to form an ideal solution. The vapour pressure of solution will be
Given:
$\text{Vapour pressure of pure A,}\left(p_A^{\circ }\right)=180mmHg$
$\text{Vaour pressure of pure B,}\left(p_B^{\circ }\right)=60mmHg$

• A. $145mmHg$
• B. $240mmHg$
• C. $120mmHg$
• D. $100mmHg$

Answer 1

The correct option is D $100mmHg$

$\text{Number of moles of A,}{n}_A=3mol\text{Number of moles of B,}{n}_B=6mol$

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

$x_A=\frac{n_A}{n_A+n_B}=\frac{3}{3+6}=\frac{1}{3}$
$x_B=\frac{2}{3}$
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.
$\text{By Raoult's law,}$
$p_A=x_A\times p_A^o$
$p_B=x_B\times p_B^o$
$P_{total}=x_A\times p_A^o+x_B\times p_B^o=180\times \frac{1}{3}+60\times \frac{2}{3}=60+40=100mmHg$