CBSE Class 12 · 2017 · SolutionsShort Answer

A 4% solution (w/w) of sucrose (M = 342 g mol

^{- 1}

) in water has a freezing point of 271.15 K. Calculate the freezing point of a 5% glucose (M = 180 g mol

^{- 1}

) solution in water. (Given: freezing point of pure water = 273.15 K)

Answer: T_f ≈ 268.35 K (*)

Solution

Sucrose solution (4 g per 100 g solution → 96 g water):

m_{s u c} = \frac{4/342}{0.096} = 0.1218 mol kg^{- 1}

Its depression =

273.15 - 271.15 = 2.00

K, so the effective constant:

K_{f} = \frac{2.00}{0.1218} = 16.42 K kg mol^{- 1}

Glucose solution (5 g per 100 g solution → 95 g water):

m_{g l u} = \frac{5/180}{0.095} = 0.2924 mol kg^{- 1}

Depression:

Δ T_{f} = 16.42 \times 0.2924 = 4.80 K

T_{f} = 273.15 - 4.80 = 268.35 K

(*) K

_{f}

implied by the question's own data is used, as intended.

CBSE Class 12 Board · 2017 · Subjective