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Question: Answered & Verified by Expert
3.0 molal aqueous solution of an electrolyte A2B3 is 50 % ionised. The boiling point of the solution at 1 atm is: KbH2O=0.52 Kkgmol-1
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Options:
  • A 274.76 K
  • B 377.68 K
  • C 374.68 K
  • D 104.68 K
Solution:
2669 Upvotes Verified Answer
The correct answer is: 377.68 K
A2B3(aq)2A3+(aq)+3B2-(aq)
n=5
ΔTb=i.Kb.m
=[1+(n-1)α]Kb.m.
=[1+4α]Kb.m.
=[1+4(0.5)]0.52×3
=(1+2)(0.52)×3
=4.68
Ab=373+4.68=377.68K

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