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A 4% solution (w/w) of sucrose (M = 342 g mol^-1) in water has freezing point of 271.15K. Calculate the freezing point of 5% glucose (M = $180 g mol^{-1}$ ) in water.

(Given: Frezzing point of water = 273.15K)

Answers (1)

$\Delta T_{f}= K_{f}\times m$

$= 16.41 \times \frac{5}{180}\times \frac{1000}{45}=4.8K$

$\Delta T_{f}= T^{\circ}_{f}- T_{f}$

$4.8=273.15-T_{f}$

$T_{f}-268.35K$

Posted by

Sumit Saini

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