Question

Vapour pressure of pure water at 27°C is 3000 k Pa. By dissolving 5g of a non volatile molecular
solid in 100g of water the vapour pressure is decreased to 2985 k Pa. What is the molecular weight
of solule?​

Answer 1

Answer:

Mole fraction of solute, x

2

=

p

1

∗

−Δp

=

3000

3000−2985

=0.005

Now, x

2

=

M

1

m

1

+

M

2

m

2

m

2

/M

2

≈

m

1

/M

1

m

2

/M

2

∴0.005=

100/18

5/M

2

⇒M

2

=180gmol

−1

∴ The molar mass of solute is 180gmol

−1

.