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Question

# 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 is 46 g mol−1). Among the following, the option representing change in the freezing point is:

A
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B
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C
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D
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Solution

## The correct option is C Depression in freezing point of ethanol and water solution: ΔTf=iKfm =1×2×34.5/46500/1000=3K So the freezing point of the solution is 270 K In option b and d, the freezing point shown is 271 K so both are incorrect Between option a and c the vapour pressure of the solution is decreasing with temperature after the freezing point in option a so it is also incorrect while option c seems to be the only possible curve for the given conditions.

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