$1.02$ g of urea when dissolved in $98.5$ g of certain solvent decreases its freezing point by $0.211 K . 1.60 g$ of an unknown compound when dissolved in $86.0$ g of certain solvent depresses the freezing point by $0.34$ K. Calculate the molar mass of the unknown compound. (Urea $N H_{2} C O N H_{2}, N = 14, C = 12, O = 16, H = 1$ )

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Solution

Amount of urea = $1.02$ g
no. of mole = $1.02 / 60$
Molar mass of urea $(N H_{2} C O N H_{2}) = 60$
Molality = $\frac{1.02}{60} \times \frac{1000}{98.5} = 0.17259$
$Δ T_{f} = k_{f} m \Rightarrow 0.211 = k_{f} \times 0.1759$ ...(i)
Let molar mass of unknown compound = M
$Δ T_{f} = k_{f} m$
$Δ T_{f} = k_{d} \times \frac{1.60}{M} . \frac{1000}{80.6}$
$0.34 = k_{f} \times \frac{19.85}{M}$ ...(ii)
divide equation (i) by (ii)
$\frac{0.211}{0.34} = \frac{0.1759}{19.85} \times M$
$M = \frac{0.211 \times 19.85}{0.34 \times 17.59} = 70 g m$

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