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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 mol1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). (Molecular weight of ethanol is 46 g mol1). 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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