The vapour pressure of a solution of a non-volatile electrolyte (A) in a solvent (B) is 95 % of the vapour pressure of the solvent at the same temperature. If MB= 0.3 M A , where M B and M A are molar masses of B and A respectively, the mass ratio of the solvent and solute are:

Question

The vapour pressure of a solution of a non-volatile electrolyte (A) in a solvent (B) is 95 % of the vapour pressure of the solvent at the same temperature. If MB= 0.3 M A , where M B and M A are molar masses of B and A respectively, the mass ratio of the solvent and solute are: (A) 0.15 (B) 5.7 (C) 0.2 (D) 4.0

Tardigrade Admin · Accepted Answer

(Po-Ps/Ps)=(n/N)=(w × M/m × W) or (P°-(95/100) P°/(95)100 P°) =(w × 0.3 m/W × m)(M=0.3 m) or (W/w)=5.7

Q. The vapour pressure of a solution of a non-volatile electrolyte (A) in a solvent (B) is 95% of the vapour pressure of the solvent at the same temperature. If MB​= 0.3MA​, where MB​ and MA​ are molar masses of B and A respectively, the mass ratio of the solvent and solute are:

0.15

5.7

0.2

4.0

Ps​Po−Ps​​=Nn​=m×Ww×M​ or 10095​P∘P∘−10095​P∘​ =W×mw×0.3m​(M=0.3m) or wW​=5.7

$\frac{P ^{o} - P _{s}}{P _{s}} = \frac{n}{N} = \frac{w \times M}{m \times W}$

or $\frac{P ^{\circ} - \frac{95}{100} P ^{\circ}}{\frac{95}{100} P ^{\circ}}$

$= \frac{w \times 0.3 m}{W \times m} (M = 0.3 m)$

or $\frac{W}{w} = 5.7$