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 m o

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 $m o l^{- 1}$ . 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

Options:

Answer:

A

Solution:

As when temperature increase, vapour pressure increases so C and D options get rejected.
$Δ T_{f} = K_{f} \times m$
$273 - T_{f}^{'} = 2 \times \frac{\frac{34.5}{46}}{0.5}$
So $T_{f}^{'} = 270 K$

TestHub · Accepted Answer

As when temperature increase, vapour pressure increases so C and D options get rejected. Δ T f = K f × m 273 - T f ′ = 2 × 34.5 46 0.5 So T f ′ = 270 K

Chemistry - Liquid Solution Question with Solution | TestHub

Options:

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