QJEEAdv17Che565: JEE Advanced 2017 Chemistry - Class 12 JEEAdv17Che565 Solution: Pure water freezes at 273 K and 1 bar. The addition of 34.5 g…

Question

Verified Solution • 2025–26 Session moderate

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

kg

m o l^{- 1} .

The figures shown below represent plots of vapor 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

A https://ik.imagekit.io/novatra/photos/d625679e1245236887b2403fcaac4db6ad53cf4be3afb11aaf0d3073ee369849.webp
B https://ik.imagekit.io/novatra/photos/3332dcb403d5f392eabbba815faebb6584458ac146294afa82dadb1ce5f4108a.webp
C https://ik.imagekit.io/novatra/photos/197d31d662b6c9ad6cce33407c42c7c4b6c2bd53535d5b460d53e0eb145ea74a.webp
D https://ik.imagekit.io/novatra/photos/797173200a6958a4b59c506fe7bf9387a5dd1849ad2f936447f003efc3cc21bd.webp

Accepted Answer

Depression in freezing point is given by $Δ$ T_f = K_fm

where $m = \frac{1000 \times w_{s o l u t e}}{M_{s o l u t e} \times w_{s o l v e n t}}$

Substituting the values, we get

$Δ T_{f} = 2 \times \frac{1000 \times 34.5}{46 \times 500} = 3 K$

Therefore, T_f = 270 K and $T_{f}^{o}$ = 273 K. Hence, option (c) indicates correct representation of the graph.

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