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Q.

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as $2 K k g m o l^{- 1}$ . The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is $46 g m o l^{- 1}$ ]
Among the following, the option representing change in the freezing point is

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a

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b

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c

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d

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answer is C.

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Detailed Solution

$\begin{array}{l} Δ T_{f} = K_{f} \cdot m \\ 273 - T_{f} = 2 \times \frac{34.5}{46} \times \frac{1000}{500} \\ T_{f} = 273 - 3 = 270 K \end{array}$

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Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol−1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is46 g mol−1 ]Among the following, the option representing change in the freezing point is