The freezing point (in `""^(@)C`) of solution containing `0.1` g of `K_(3) [ Fe(CN)_(6)]` (mol.wt. `329`) in `100` g of water (`K_(f)` `= 1.86` K kg `mol^(-1)`) is

Options

A. `-2.3 xx 10^(-2)`
B. `-5.7 xx 10^(-2)`
C. `5.7 xx 10^(-3)`
D. `-1.2 xx 10^(-2)`

Answer

A

Explanation

`k_(3) [Fe(CN)_(6)] hArr 3k^(+) + [Fe(CN)_(6)]^(3-)`
Here ` i = 4`
`Delta T_(f) = (ixx K_(f) xx 1000 xx w_(2))/(w_(1) xx M_(2))`
` = (4 xx 1.86 xx 1000 xx 0.1)/(100 xx 329) = 2.3 xx 10^(-2)`
Freezing point `= 0-2.3 xx 10^(-2)`
` = -2.3 xx 10^(-2)""^(@)C`.