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0.004 M Na2S04 aqueous solution is isotonic with 0.01 M glucose solution at 300 K. Thus, degree of dissociation of Na2S04 is
  • a)
    75 %
  • b)
    50 %
  • c)
    25 %
  • d)
    85 %
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
0.004 M Na2S04 aqueous solution is isotonic with 0.01 M glucose soluti...
i (van’t Hoff factor) = 1 + (y — 1) x = (1 + 2 x)
∴ π, (osmotic pressure) = MRTi = 0.004 x R x 300 x (1 + 2x)
Glucose is a non-electrolyte.
Hence,
i = 1
π2 = mRTi = 0.01 x R x 300 x 1 Two solutions are isotonic, hence
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0.004 M Na2S04 aqueous solution is isotonic with 0.01 M glucose solution at 300 K. Thus, degree of dissociation of Na2S04 isa)75 %b)50 %c)25 %d)85 %Correct answer is option 'A'. Can you explain this answer?
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