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 $2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is $46 \mathrm{~g} \mathrm{~mol}^{-1}$] Among the following, the option representing change in the freezing point is

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Solutions

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)(see structure)
(B)(see structure)
(C)(see structure)
✓ Correct
(D)(see structure)

Answer: (C)

Solution

The freezing point depression is calculated using

Δ T_{f} = K_{f} \times m

.

Molality

m = \frac{34.5 g}{46 g/mol} \times \frac{1000}{500 g} = 1.5 mol/kg

Δ T_{f} = 2 \times 1.5 = 3

K

Freezing point of solution =

273 - 3 = 270

K

The correct plot shows the freezing point shifted from 273 K to 270 K, with the vapor pressure curves for pure water and solution intersecting the solid line at these respective temperatures.

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option

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JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Solutions

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