1. 3 g of a salt of molecular weight 30 is dissolved in 250 g of water. The molality of the solution is _____________ . (JEE 1983)

Answer; 0.4 mol kg^-
Molality (m) = number of moles of solute/kilograms of solvent
0.1 mol salts dissolved in .25 kg of water. So m = 0.1/.25 = 0.4 mol/kg

2. Given that ΔT_f is the depression in freezing point of the solvent in a solution of a non-volatile solute of molality, m, te quantity lim m→0 (ΔT_f/m) is equal to _________________ . (JEE 1994)

Answer: K_f