Question

A solution is prepared by mixing 4 mole of liquid A, 6 mole of liquid B, 0.4 mole of non-volatile solute C and 0.2 moles of non-volatile solute D. If C undergoes 40% dimerisation while D undergoes 30% binary dissociation, then the vapour pressure of solution is :

(Given, vapour pressure of pure solvent A and B are 500 mm and 600 mm of Hg respectively)

• A. $490.3$ mm
• B. $500.3$ mm
• C. $529.3$ mm
• D. $539.3$ mm

Answer 1

The correct option is C $529.3$ mm

The moles of C after the association is $0.32=0.4-\frac{0.4\times 0.4}{2}$.

The moles of D after dissociation is $0.26=0.2+0.2\times 0.3$.

$\therefore molefractionofA=\frac{4}{10+0.32+0.26}$

$molefractionofB=\frac{6}{10+0.32+0.26}$

$\therefore Vapourpressureofsolution=P_A^0{X}_A+P_B^0{X}_B=500\times \frac{4}{10.58}+600\times \frac{6}{10.58}=529.3$ mm of Hg