For a solution of 0.89 g of mercurous chloride in 50 g of `HgCl_(2)(l)` the freezing point depression id `1.24^(@)`C. `K_(f)` for `HgCl_(2)` is 34.3 . What is the state of mercurous chloride in `HgCL_(2)`? (Hg-200, Cl-35.5).

Options

A. as `Hg_2Cl_2` molecules
B. as `HgCl` molecules
C. as `Hg^+` and `Cl^(-)` ions
D. as `Hg_2^2+ and Cl^(-)` ions

Answer

A

Explanation

`1.24=34.3[((0.849)/M)/0.050]implies M=469.68`