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NTA Abhyas JEE Main2020ChemistrySolutionsPractice

The freezing point of a 0.08 molal aqueous solution of N a H S O 4 is - 0.372 ° C . The dissociation constant for the following reaction is ( K f for H 2 O = 1.86 K kg mo l - 1 ) H S O 4 - ⇌ H + + S O 4 2 -

Options

  1. A0.04
  2. B0.02
  3. C0.01
  4. D0.2

Correct answer

A. 0.04

Step-by-step solution

N a H S O 4 → N a + 0.08 + H S O 4 - 0.08 H S O 4 - 0.08 ( 1 - α ) ⇌ H + 0.08 α + S O 4 2 - 0.08 α i = 0.08 + 0.08 1 - α + 0.08 α + 0.08 α 0.08 = 2 + α Δ T = i × K f × m 0.372 = i × 1.86 × 0.08 i = 2.5 So 2 + α = 2.5 α = 0.5 Dissociation constant, K = 0.08 α × 0.08 α 0.08 1 - α = 0.08 × 0.5 × 0.08 × 0.5 0.08 × 0.5 = 0.04

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