Question

The freezing point (in °C) of solution containing 0.18 g of glucose in 100 g water is (Kf = 1.86 K kg mol–1) is ?

A)–3.7 × 10–2
B) –5.7 × 10–2
C) –1.86 × 10–2
D) –1.2 × 10–2

Answer 1

∆Tf = i kf.m

∆Tf = 1.86 × (0.18/ 180 )× 1000/ 100

∆Tf = 1.86 × 0.01

∆Tf = 0.0186 K

∆Tf = Tb - Ts

Ts = 0 ( pure water temp) - ∆Tf

Ts = 0 - 0.0186

Ts = - 0.0186 or –1.86 × 10^(–2) ° C

Correct option is C) .