Van't hoff factor of Hg2Cl2 in its Aqueous solution will be (it is 80% ionised in the solution) ? (1) 1.6 (2)2.6 (3)3.6 (4) 4.6

I get #i = 2.60#, if we pretend #"Hg"_2"Cl"_2# is a strong electrolyte in water... but you must keep in mind that it is also very insoluble in water, and that dissociation is distinct from dissolution. Here we apparently have a poorly soluble strong electrolyte.

Well, to do this, we must already know that the cation is not #"Hg"^(+)#, but #"Hg"_2^(2+)#. Suppose the starting concentration was #c#. Then:

#"Hg"_2"Cl"_2(aq) rightleftharpoons "Hg"_2^(2+)(aq) + 2"Cl"^(-)(aq)#

#"I"" "" "c" "" "" "" "" "0" "" "" "" "0#
#"C"" "-alphac" "" "" "+alphac" "" "+2alphac#
#"E"" "(1-alpha)c" "" "" "alphac" "" "" "2alphac#

where #0 < alpha < 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 #K_(sp)# would have been written:

#K_(sp) = ["Hg"_2^(2+)]["Cl"^(-)]^2 = s(2s)^2 = 4s^3#

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

#["Hg"_2"Cl"_2]_(eq) + ["Hg"_2^(2+)]_(eq) + ["Cl"^(-)]_(eq)#

#= (1-alpha)c + alphac + 2alphac#

#= (1 + 2alpha)c#

The van't Hoff factor is defined to be:

#i = (["dissociated species concentration"])/(["undissociated concentration"])#

From this, we find:

#color(blue)(i) = ((1 + 2alpha)c)/c = 1 + 2alpha#

#= 1.00 + 2 cdot 0.80#

#= color(blue)(2.60)#