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Solutions question

2003 · Shift 0 · Q31
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Solutions question

2003 · Shift 0 · Q31

JEE MainChemistrySolutionsMCQ+4 / −1
In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking kf for water as 1.85, the freezing point of the solution will be nearest to
  1. A
    -0.360oC
  2. B
    -0.260oC
  3. C
    +0.480oC
  4. D
    -0.480oC
View written solutionFree

Correct answer: D

  1. Use freezing point depression formula

For a solution, ΔTf=iKfm\Delta T_f = i K_f mΔTf​=iKf​m where:

  • iii = van’t Hoff factor
  • Kf=1.85K_f = 1.85Kf​=1.85
  • m=0.2m = 0.2m=0.2
  1. Find van’t Hoff factor for weak acid HXHXHX

The acid ionizes as: HX⇌H++X−HX \rightleftharpoons H^+ + X^-HX⇌H++X−

If degree of ionization is α=0.3\alpha = 0.3α=0.3, then starting with 1 mole of HXHXHX:

  • undissociated = 1−α1-\alpha1−α
  • ions formed = α+α=2α\alpha + \alpha = 2\alphaα+α=2α

So total moles in solution: i=(1−α)+2α=1+αi = (1-\alpha) + 2\alpha = 1+\alphai=(1−α)+2α=1+α i=1+0.3=1.3i = 1 + 0.3 = 1.3i=1+0.3=1.3

  1. Calculate depression in freezing point

ΔTf=1.3×1.85×0.2\Delta T_f = 1.3 \times 1.85 \times 0.2ΔTf​=1.3×1.85×0.2

First, 1.85×0.2=0.371.85 \times 0.2 = 0.371.85×0.2=0.37

Then, 0.37×1.3=0.4810.37 \times 1.3 = 0.4810.37×1.3=0.481

Thus, ΔTf≈0.48∘C\Delta T_f \approx 0.48^\circ CΔTf​≈0.48∘C

  1. Find actual freezing point

Since pure water freezes at 0∘C0^\circ C0∘C, solution freezing point is: 0−0.48=−0.48∘C0 - 0.48 = -0.48^\circ C0−0.48=−0.48∘C

  1. Match with options

The nearest value is: −0.480∘C\boxed{-0.480^\circ C}−0.480∘C​ So, Option D is correct.

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