JEE Advanced 2017 · Paper 2 · Chemistry Q1

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ChemistrySingle correct+3 / -1Question 19 of 54

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 $2\ \mathrm{K\,kg\,mol^{-1}}$ . The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is $46\ \mathrm{g\,mol^{-1}}$ ]

Among the following, the option representing change in the freezing point is

V.P./bar against T/K. The ice curve rises to meet the water curve at 273 K (V.P. = 1 bar) and meets the water + ethanol curve at 270 K; to the right of these points both liquid curves fall with temperature.

V.P./bar against T/K, all curves rising. The ice curve meets the water curve at 273 K and the lower water + ethanol curve at 271 K.

V.P./bar against T/K, all curves rising. The ice curve meets the water curve at 273 K and the lower water + ethanol curve at 270 K.

V.P./bar against T/K. The ice curve rises to meet the water curve at 273 K and the water + ethanol curve at 271 K; both liquid curves fall with temperature beyond these points.

Answer and worked solution

Official answer: C

Depression.
$m=\frac{34.5/46}{0.5}=1.5\ \mathrm{mol\,kg^{-1}},\qquad\Delta T_f=2\times1.5=3\text{ K}.$
The solution freezes at 270 K.

The right plot. Vapour pressure rises with temperature for ice, water and the solution. The solution's curve lies below pure water's. The ice curve meets the water curve at 273 K and the solution's curve at 270 K. That is plot (C). (A) and (D) show liquid vapour pressure falling with temperature; (B) puts the solution's freezing point at 271 K.

(C)

Topics:Colligative properties Raoult's law and vapour pressure

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JEE Advanced 2017 · Paper 2, Chemistry question 1