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Standard XII

Chemistry

Introduction

A 4 % solutio...

Question

A 4% solution(w/w) of sucrose (M = 342 $g m o l^{- 1})$ in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 $g m o l^{- 1})$ in water. (Given: Freezing point of pure water = 273.15 K)

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Solution

Molality of 4% solution of sucrose:
let assume that 4 grams sucrose in 96 grams of water then molality of solution
$m = \frac{4 \times 1000}{342 \times 96} (1)$
$△ (T_{f})_{s u c r o s e (a q .)} = i \times m \times k_{f}$
$273.15 - 271.15 = 1 \times \frac{4 \times 1000}{342 \times 96} \times k_{f} (2)$
the molal freezing point depression constant of the solvent which is water for both solution , from eqn. 2 the value of $k_{f} = \frac{2 \times 342 \times 96}{4 \times 1000} (3)$

For 5% solution of glucose:
$△ (T_{f})_{g l u c o s e (a q .)} = i \times \frac{5 \times 1000}{180 \times 95} \times k_{f}$
after putting value of $k_{f}$ from eqn. (3)
$△ (T_{f})_{g l u c o s e (a q .)} = 4.8$
$273.15 - T_{f} = 4.8$
$T_{f} = 268.35 K$

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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 5% glucose (M = 180 g mol–1) in water. (Given: Freezing point of pure water = 273.15 K)

A 4% solution(w/w) of sucrose (M = 342 $g m o l^{- 1})$ in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 $g m o l^{- 1})$ in water. (Given: Freezing point of pure water = 273.15 K)