The freezing point of an aqueous solution of `0.1` m `Hg_(2)Cl_(2)` will be :
(If `Hg_(2)Cl_(2)` is `80%` ionised in the solution to give `Hg_(2)^(2+)` and `Cl^(-)` :

If Hg_(2)Cl_(2) is 100% ionised, then find out its van't Hoff factor.

The Van't Hoff factor of Hg_(2)Cl_(2) in its aqueous solution will be ( Hg_(2)Cl_(2) is 80% ionized in the solution) a. 1.6 , b. 2.6 ,c. 3.6 ,d. 4.6

The Van't Hoff factor of Hg_(2)Cl_(2) in its aqueous solution will be ( Hg_(2)Cl_(2) is 80% ionized in the solution) a. 1.6 , b. 2.6 ,c. 3.6 ,d. 4.6

The Van't Hoff factor of Hg_(2)Cl_(2) in its aqueous solution will be ( Hg_(2)Cl_(2) is 80% ionized in the solution) a. 1.6 , b. 2.6 ,c. 3.6 ,d. 4.6

The freezing point of an aqueous solution of KCN containing 0.1892 mol kg^(-1) was -0.704^(circ)C . On adding 0.45 mole of Hg(CN)_(2) , the freezing point of the solution was =0.620^(circ)C . If whole of Hg(CN)_(2) is used in complex formation according to the equation, Hg(CN)_(2)+mKCN rarrK_(m)[Hg(CN)_(m+2)] what is the formula of the complex ? Assume [Hg(CN)_(m+2)]^(m-) is not ionised and the complex molecule is 100% ionised. (K_(f)(H_(2)O) is 1.86 kg mol^(-1) K .)

Calomel (Hg_(2)Cl_(2)) on reaction with ammonium hydroxide gives