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. `-2^(@)C`
B. `-3^(@)C`
C. `-1.5^(@)C`
D. `-1.66^(@)C`

Answer

B

Explanation

`Delta T_(f) = iK_(f)m`
For NaCl , `Delta T_(f) = 2, I = 2`
`2 = 2 xx k_(f) xx m`
For `BaCl_(2), Delta T_(f) = ? , I = 3`
`Delta T_(f) = 3 xx K_(f) xx m`
Since `K_(f)` and m are same (equimolal solution) .
`(Delta T_(f))/(2) = (3)/(2) or Delta T_(f) = 3 `
`therefore` Freezing point of `BaCl_(2)` solution `= 0-3 =-3^(@)C`