Question

the vapour pressure of pure water at 31 degree centigrade is 32mm hg. when 2 grams of a non volatile solute was dissolved in 36 grams of water, the vapor pressure was found to be 31.5 mm hg.calculate molar mass of the solute

Answer 1

Explanation:

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

Hence, the correct option is D