Question

11. A beaker of a liquid with a vapour pressure of 350 torr at 25°C is set alongside a beaker of water (Vapour pressure of 23.76 torr) and both are...

Answer 1

ANSWER

Let M be the molecular weight of solute.

Number of moles of solute =  

M

5.40

​  

 

Number of moles of water =  

18

90.0

​  

=5

Mole fraction of solute =  

M

5.40

​  

+5

M

5.40

​  

 

​  

=  

5.40+5M

5.40

​  

 

Relative lowering in the vapour pressure =  

P  

0

 

P  

0

−P

​  

=  

23.76

23.76−23.32

​  

=0.01852

The relative lowering in the vapour pressure of the solution is equal to the mole fraction of solute.  

0.01852=  

5.40+5M

5.40

​  

 

⟹5.40+5M=291.6

5M=286.2

⟹M=57.24 g/mol

Hope it will help you!

Explanation: