Question

24. If 30 g of a solute of molecular weight 150 is dissolved
in 250 g of benzene, what will be the boiling point of the
resulting solution under atmospheric pressure?​

Answer 1

Given:

w = 30 gm

M. wt = 150 gm/mol

W = 250 gm

To Find:

The boiling point of the resulting solution.

Calculation:

- We know that for benzene:

T° = 353.3 K

Kb = 2.53

- Putting this in the formula of elevation in boiling point, we get:

ΔTb = Kb × m

⇒ (Tb - T°) = Kb × (w × 1000)/(M.wt × W)

⇒ (Tb - 353.3) = 2.53 × (30 × 1000)/(150 × 250)

⇒ (Tb - 353.3) = 2.53 × (30 × 1000)/(150 × 250)

⇒ Tb = (2.53 × 0.8) + 353.3

⇒ Tb = 2.024 + 353.3

⇒ Tb = 255.324 K

- So, the boiling point of the  resulting solution under atmospheric pressure is 255.324 K