$1.0 \times 10^{- 3}$ kg of urea when dissolved in $0.0985$ kg of a solvent, decreases freezing point of the solvent by $0.211$ k $1.6 \times 10^{- 3}$ kg of another non-electrolyte solute when dissolved of $0.086$ kg of the same solvent depresses the freezing point by $0.34$ K. Calculate the molar mass of the another solute.(Given molar mass of urea $= 60$ )

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Solution

Given: Mass of solute, Urea $= W_{2} = 1.0 \times 10 - 3 k g = 1 g$
Mass of solvent $= W_{2} = 0.0985 k g = 98.5 g$
$∴$ $Δ T = 0.211$ k
Molar mass of urea $(N H_{2} C O N H_{2}) = M_{2} = 60 g / m o l$ .
Mass of unknown substance $= W_{2} = 1.6 \times 10^{- 3} k g = 1.6 g$
Mass of a Solvent $= W_{1}^{'} = 0.086$ kg
$∴ Δ T_{F} = 0.34$ k
Molar mass of unknown substance $= M_{2}^{'} = ?$
This problem will be done in two parts
(i) To calculate molal depression constant $^{'} k_{f}^{'}$ ,
For urea solution-
$Δ T_{F} = K_{f} \times \frac{W_{2} \times 1000}{W_{1} \times M_{2}}$
$∴ K_{f} = \frac{Δ T_{F} \times W_{1} \times M_{2}}{W_{2} \times 1000}$
$= \frac{0.211 \times 98.5 \times 60}{1 \times 1000}$
$= \frac{1247.01}{1000} = 1.24701$
$∴ K_{F} = 1.2 g / m o l$ .
(ii) To calculate molar mass, $M_{2}$
For unknown solution
$M_{2}^{'} = K_{F} \times \frac{W_{2}^{'} \times 1000}{W_{1}^{'} \times 0.34}$
$∴ K_{F} = \frac{1.2 \times 1.6 \times 1000}{86 \times 0.34} = \frac{1920}{29.24}$
$∴ M_{2}^{'} = 65.66 g / m o l$ .
Hence, the molar mass of another solution is $65.66 g / m o l$ .

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