The freezing point (in `.^(@)C)` of a solution containing `0.1g` 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. `-2.3xx 10^(-2)`
B. `-5.7 xx 10^(-2)`
C. `-5.7 xx 10^(-3)`
D. `-1.2 xx 10^(-2)`

Answer

A

Explanation

`DeltaT_f = ixxK_f xxm`
Where m = Molality of the solution
(i.e. Number of moles of solute per 1000 g of the solvent)
Here `m= 0.1/329 xx 10`
Thus `DeltaT_f = 4 xx 1.86 xx (0.1xx10)/329 = 2.3 xx 10^(-2)`
Thus, `T_f = 0 -2.3 xx 10^(-2) = -2.3 xx 10^(-2)^@C`