Question
A non-volatile solute ‘X’ completely dimerizes in water, if the temperature is below −3.72°C and the solute completely dissociates as X → Y + Z, if the temperature is above 100.26°C. In between these two temperatures (including both temperatures), the solute is neither dissociated nor associated. One mole of ‘X’ is dissolved in 1.0 kg water (Kb = 0.52 K-kg/mol, Kf = 1.86 K-kg/ mol). Identify the incorrect information related with the solution.
Options
Solution
(a) $\Delta T_{f} = K_{f}.m = 1.86 \times 1 = 1.86^{o}C$
∴ F.P. of solution = – 1.86° C
(b) $\Delta T_{b} = K_{b}.m = 0.52 \times 1 = 0.52^{o}C$
∴ B.P. of solution should be 100.52° C. As the solute dissociates completely above 100.26° C, its actual $\Delta T_{b} = 0.52 \times 2 = 1.04^{o}C$ and hence, B.P. = 101.04° C.
(c) $\underset{\left( \text{as solute dimerizes} \right)}{\Delta T_{f} = K_{f}.m \Rightarrow}7.44 = 1.86 \times \frac{1/2}{m_{\text{solvent}}}$
∴ mSolvent left = 0.125 kg
∴ Percentage of water separated as ice = (1 – 0.125) × 100 = 87.5%
(d) $\underset{\left( \text{Complete dissociation} \right)}{\Delta T_{b} = K_{b}.m \Rightarrow 2.08 = 0.52 \times \frac{2}{m_{\text{solvent}}}}$
∴ mSolvent left = 0.5 kg
Percentage of water evaporated = (1 – 0.5) × 100 = 50%
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