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Solutions question

2022 · 26 Jul · Shift 2 · Q15
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Solutions question

2022 · 26 Jul · Shift 2 · Q15

JEE MainChemistrySolutionsNumerical+4 / −1
The elevation in boiling point for 1 molal solution of non-volatile solute A is 3 K3 \mathrm{~K}3 K. The depression in freezing point for 2 molal solution of A\mathrm{A}A in the same solvent is 6 KKK. The ratio of KbK_{b}Kb​ and KfK_{f}Kf​ i.e., Kb/KfK_{b} / K_{f}Kb​/Kf​ is 1:X1: X1:X. The value of XXX is [nearest integer]
Numerical answer
View written solutionFree

Correct answer: 1

  1. Use colligative property formulas

For a non-volatile solute:

ΔTb=Kbm\Delta T_b = K_b mΔTb​=Kb​m ΔTf=Kfm\Delta T_f = K_f mΔTf​=Kf​m

where mmm is molality.

  1. Find KbK_bKb​ from the first data

Given for a 111 molal solution:

ΔTb=3 K\Delta T_b = 3\,\text{K}ΔTb​=3K m=1m = 1m=1

So,

Kb=ΔTbm=31=3K_b = \frac{\Delta T_b}{m} = \frac{3}{1} = 3Kb​=mΔTb​​=13​=3

  1. Find KfK_fKf​ from the second data

Given for a 222 molal solution:

ΔTf=6 K\Delta T_f = 6\,\text{K}ΔTf​=6K m=2m = 2m=2

So,

Kf=ΔTfm=62=3K_f = \frac{\Delta T_f}{m} = \frac{6}{2} = 3Kf​=mΔTf​​=26​=3

  1. Calculate the ratio Kb/KfK_b/K_fKb​/Kf​

KbKf=33=1\frac{K_b}{K_f} = \frac{3}{3} = 1Kf​Kb​​=33​=1

Given that

Kb:Kf=1:XK_b:K_f = 1:XKb​:Kf​=1:X

So,

1:11:11:1

Hence,

X=1X = 1X=1

  1. Compare with stored correct answer

Stored correct answer = 111

Our derived answer = 111

So the answer agrees.

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