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Solutions question

2021 · 18 Mar · Shift 2 · Q22
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Solutions question

2021 · 18 Mar · Shift 2 · Q22

JEE MainChemistrySolutionsNumerical+4 / −1
A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52 ∘^\circ∘ C. The percentage association of A is ‾\underline{\hspace{2cm}}​. (Round off to the Nearest Integer). [Use : Kb for water = 0.52 K kg mol −-− 1 Boiling point of water = 100 ∘^\circ∘ C]
Numerical answer
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Correct answer: 100

  1. Use boiling point elevation relation

For a solution, ΔTb=iKbm\Delta T_b = i K_b mΔTb​=iKb​m where:

  • ΔTb\Delta T_bΔTb​ = elevation in boiling point
  • iii = van't Hoff factor
  • Kb=0.52 K kg mol−1K_b = 0.52\,\text{K kg mol}^{-1}Kb​=0.52K kg mol−1
  • m=2m = 2m=2

Given boiling point of solution =100.52∘C= 100.52^\circ\text{C}=100.52∘C and pure water =100∘C= 100^\circ\text{C}=100∘C, ΔTb=100.52−100=0.52 K\Delta T_b = 100.52 - 100 = 0.52\,\text{K}ΔTb​=100.52−100=0.52K

  1. Calculate van't Hoff factor

0.52=i×0.52×20.52 = i \times 0.52 \times 20.52=i×0.52×2 0.52=1.04i0.52 = 1.04 i0.52=1.04i i=0.521.04=0.5i = \frac{0.52}{1.04} = 0.5i=1.040.52​=0.5

  1. Relate iii to association

If solute AAA dimerizes: 2A⇌A22A \rightleftharpoons A_22A⇌A2​

Assume initially 1 mole of AAA. Let degree of association be α\alphaα.

Then:

  • moles of AAA left =1−α= 1-\alpha=1−α
  • moles of dimer formed =α2= \frac{\alpha}{2}=2α​

Total moles in solution after association: 1−α+α2=1−α21-\alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}1−α+2α​=1−2α​

Hence van't Hoff factor: i=1−α2i = 1 - \frac{\alpha}{2}i=1−2α​

Substitute i=0.5i=0.5i=0.5: 0.5=1−α20.5 = 1 - \frac{\alpha}{2}0.5=1−2α​ α2=0.5\frac{\alpha}{2} = 0.52α​=0.5 α=1\alpha = 1α=1

  1. Convert to percentage association

% association=α×100=100%\%\text{ association} = \alpha \times 100 = 100\%% association=α×100=100%

  1. Final answer

Rounded to nearest integer: 100\boxed{100}100​

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