1.

The Vapour pressure of water at 293 K is 0.0231 bar and the vapour pressure of the solution of 108.24 g of a compound in 1000 g of water at the same temperature. Calculate the molar mass of the solute.

Answer»

Vapour pressure of solvent 

pA0 = 0.0231 bar 

Vapour pressure of solution

pA0 = 0.0228 bar 

Lowering in vapour pressure 

pA0  - pA = 0.0231 - 0.0228 = 0.0003 bar 

Weight of solvent WA = 1000 g 

Weight of solute WB = 108.24 g 

Molar mass of the solvent MA = 18 

Molar mass of the solute MB = ? 

We know that 

pA˚ - pA / p0A = WB MA/WA MB

MB = WB MA PA˚/ WA (PA˚ - PA) 

= 108.24 X 0.0231 X 18/ 0.0003 X 1000

= 150 g mol-1 



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