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Question

If Hg2Cl2 is 80% ionised in the aqueous solution what will be the van't Hoff factor of Hg2Cl2 in the solution?

A
1.6
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B
2.6
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C
3.6
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D
4.6
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Solution

The correct option is B 2.6
Hg2Cl2 (aq)Hg2+2+2ClInitially: C 0 0Eqb: CCα Cα 2Cα

i=No. of moles after dissociationNo. of moles before dissociation

i=CCα+Cα+2CαCi=C+2CαC=1+2α
Given that :
α=0.8
i=1+(2×0.8)i=2.6

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