`0.01` M solution of KCl and `Ca Cl _(2)` are separately prepared in water. The freezing point of KCl is found to be `- 2 ^(@) C.` What is the freezing point of `Ca Cl _(2)` aq. Solution if it is completely ionized ?

Options

A. `- 3 ^(@) C`
B. `+ 3 ^(@) C`
C. `- 2 ^(@) C`
D. `- 4 ^(@) C`

Answer

A

Explanation

I for `KCl = 2,` i for ` Ca Cl _(2) = 3`
`Delta T _(f) prop i`
`( Delta T _(f) (KCl ))/( Delta T _(f) (Ca Cl _(2))) = (2)/(3)`
`Delta T _(1) (Ca Cl _(2)) = (3)/(2) xx 2 = 3 ^(@) C`
Freezing point of `Ca Cl _(2) = - 3 ^(@) C`