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JEE Advanced Chemistry Solutions 2021 JEE Advanced 2021 (Paper 1)

JEE Advanced Chemistry Question (2021) — Solution

Question

The boiling point of water in a 0.1 molal silver nitrate solution (solution A ) is x ^ C . To this solution A , an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y 10^ -2 ^ C . (Assume: Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely. Use: Molal elevation constant (Ebullioscopic Constant), K_ b =0.5 ~K ~kg ~mol ^ -1 ; Boiling point of pure water as 100^ C .) The value of x is ___.

Options

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

Answer

A. A

Step-by-step solution

Given for solution  A m = 0 . 1 i=  1  for  AgNO 3 = 2 K b = 0 .5   Kg   mol - 1 So  ΔT b = T b − T b ∘ = 1   K b m T b − 100 ∘ C = 2 × 0 .5 × 0 .1 T b = 100 ∘ C + 0 .1 = 100 .1 ∘ C

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