Question

What is the freezing point of solution prepared by disolving 15 g of KBr in 100 ml of water? The freezing point depression constant of water is 1.86.

Answer 1

ΔT_f=K_f×Cm

K_f(H₂O)=1.86

Cm (molality) is number of moles of solute per kilogram solvent.

ΔT_f=K_f×m

M (KBr) = 119 g/mol

n (NaCl) = 15/119 = 0.13 mol

Density of water is 1 kg/l. Threfore, mass of 100 ml = 0.1 L of water is 0.1 kg.

Сm (KBr) = 0.13/0.1=1.3 mol

ΔT_f=1.86×1.3=2.4

Therefore T_f of an aqueous solution containing 15 g of KBr in 100 ml of H₂O will be -2.4°C