Question

can someone answer this question​

Answer 1

Well, to do this, we must already know that the cation is not Hg+, but Hg2+2. Suppose the starting concentration was c. Then:

Hg2Cl2(aq)⇌Hg2+2(aq)+2Cl−(aq)

I c 0 0
C −αc +αc +2αc
E (1−α)c αc 2αc

where 0<α<1 is the fraction of dissociation, i.e. fraction of ionization.

As usual, note that the coefficient on each species carries through the changes in concentration. Note also that the Ksp would have been written:

Ksp=[Hg2+2][Cl−]2=s(2s)2=4s3

Anyways, after this, the total concentration of all mercury/chloride species becomes:

[Hg2Cl2]eq+[Hg2+2]eq+[Cl−]eq

=(1−α)c+αc+2αc

=(1+2α)c

The van't Hoff factor is defined to be:

i=[dissociated species concentration][undissociated concentration]

From this, we find:

i=(1+2α)cc=1+2α

=1.00+2⋅0.80

=2.60


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