Question

The boiling point of water (100°C) becomes 100.52°C, if 3 g of a non-volatile solute is dissolved in 200 mL of water. The molecular weight of solute is (K_b for water = 0.6 K/m)

(a) 17.3 g mol^-1 

(b) 15.4 mol^-1

(c) 12.2 g mol^-1

(d) 20.4 g mol^-1

Answer 1

Correct option (a) 17.3 g mol^-1 

Explanation:

As elevation in boiling point is given by :

ΔT_b = K_b x molality

Putting the values, we get :