Question

The vapour pressure of
benzene at 80°C is lowered
by 10 mm by dissolving 2g
of a non volatile substance
in 78g of benzene. The
vapour pressure of pure
benzene at 80°C is 750 mm.
The molecular mass of the
substance will be
15
150
1500
190​

Answer 1

Answer:

150 is the correct answer

Explanation:

hope the answer will help you

Answer 2

Given:

The lowering of vapour pressure (P°-P) = 10 mm Hg

The vapour pressure of pure benzene, P° = 750 mm Hg

The mass of solute, w = 2 gm

The mass of benzene, W = 78 gm

To Find:

The molecular mass of the  non- volatile solute.

Calculation:

- Let the molecular mass of the  non- volatile solute be M.

- The no of  moles of solute, n1 = 2/M

- The no of  moles of benzene, n2 = 78/78 = 1

- Lowering of vapour pressure = X(solute)

⇒  (P°-P)/ P° = n1 / (n1 + n2)

⇒ 10/750 = (2/M) / {(2/M) + 1}

⇒ {(2/M) + 1} × 10= (2/M) × 750

⇒ (2 + M)/ M = 150/M

⇒ M = 150 -2

⇒ M = 148 gm

- So, the molecular mass of the  non- volatile solute is 148 gm.