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Question

Chemistry Question on Colligative Properties

Pure water freezes at 273K273 \,K and 11 bar. The addition of 34.5g34.5\, g of ethanol to 500g500\, g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2Kkgmol12\,K\, kg\,mol^{-1}. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46gmol146\, g\,mol^{-1}] Among the following, the option representing change in the freezing point is

A

B

C

D

Answer

Explanation

Solution

ΔTf=Kf×m\Delta T _{ f }= K _{ f } \times m
=2×34.5×246=2 \times \frac{34.5 \times 2}{46}
=2×1.5=2 \times 1.5
=3=3
Freezing point of ethanol ++ water mixture
=2733=270=273-3=270