# (a) A 10% solution (by mass) of sucrose in water has a freezing point of 269·15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273·15 K. Given : (Molar mass of sucrose = 342 g mol-1) (Molar mass of glucose = 180 g mol-1) (b) Define the following terms : (i) Molality (m) (ii) Abnormal molar mass

(a) Molor mass of sucrose = 342g mol-1
Molar mass of glucose = 180 g mol-1
Tf  = ?
$\Delta T_{f}= K_{f}m$
$\Delta T_{f}= T_{f^{0}}-T_{f}$
$T_{f^{0}}-T_{f}= K_{f}m$
$273\cdot 15-269\cdot 15= K_{f}\times \frac{10}{342}\times \frac{1000}{100}$
$K_{f}= \frac{4\times 342}{10\times 10}$
$= 13\cdot 68Kmol^{-1}$
$273\cdot 15-T_{f} = 13\cdot 68\times \frac{10}{180}\times \frac{1000}{100}$
$273\cdot 15-7\cdot 6= T_{f}$
$T_{f}= 273\cdot 15-7\cdot 6=265\cdot 55 K$
(b) (i) Molality (m)
as the number of moles of solute present per kg of solvent unit of m is mol kg-1 or molal
(ii) Abnormal molar mass - If the molar mass of the solute calculated by using the colligative properly is formed to be different than its normal molar mass than the solute is said to exhibit/possess abnormal molar mass.

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