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JEE Advanced Chemistry Solutions 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Chemistry Question (2022) — Solution

Question

An aqueous solution is prepared by dissolving 0 . 1  mole of an ionic salt in 1 . 8   kg of water at 35 ° C . The salt remains 90 % dissociated in the solution. The vapour pressure of the solution is 59 . 724   mm of Hg . Vapor pressure of water at 35 ° C is 60 . 000   mm of Hg . The number of ions present per formula unit of the ionic salt is _____.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Number of ions present per formula unit of ionic salt = x Van 't Hoff factor i = 1 + n - 1 α     =   1 + x - 1 × 0 . 9   =   0 . 1 + 0 . 9 x (Assuming 90 % dissociation) ∴ Relative lowering in vapour pressure = i × Mole fraction of solute ⇒ 60 - 59 . 724 60 = i × 0 . 1 1800 18 + 0 . 1 ⇒    0 . 0046 = i × 0 . 1 100 + 0 . 1 0 . 0046 ≈ 0 . 9 x + 0 . 1 × 0 . 1 100 ⇒ 0 . 9 x + 0 . 1 = 4 . 6 ⇒    x = 4 . 5 0 . 9 = 5 x = 5

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