0.01 M solution of KCl and `CaCl_(2)` are separately prepared in water. The freezing point of KCl is found to be `-2^(@)C`. What is the freezing point of `CaCl_(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 `BaCl_(2) = 3`
`because DeltaT_(f)propi`
`(DeltaT_(f)(KCl))/(DeltaT_(f)(BaCl_(2)))=(2)/(3)`
`DeltaT_(f)(BaCl_(2))=(3)/(2)xx2=3^(@)C`
`therefore` Freezing point of `BaCl_(2)=-3^(@)C`.