Question

Which of the following solution has a higher freezing point ? 0.05 M Al2(SO4)3, 0.1 M K3[Fe(CN)6] justify.

Answer 1

Al2(SO4)3 is the correct answer

Answer 2

Answer : The correct option is, (b) 0.1 M $K_3[Fe(CN)_6]$

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_f=i\times k_f\times m$

where,

$\DeltaT_f$ = change in freezing point

$k_f$ = freezing point constant

m = molality

i = Van't Hoff factor

As we know that the boiling point depends on the molality and the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) The dissociation of 0.05 M $Al_2(SO_4)_3$ will be,

$Al_2(SO_4)_3\rightarrow 2Al^{3+}+3SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $2Al^{3+}+3SO_4^{2-}=(2\times 0.05)+(3\times 0.05)=0.25M$

(b) The dissociation of 0.1 M $K_3[Fe(CN)_6]$ will be,

$K_3[Fe(CN)_6]\rightarrow 3K^++[Fe(CN)_6]$

So, Van't Hoff factor = Number of solute particles = $3K^++[Fe(CN)_6]=(3\times 0.1)+(1\times 0.1)=0.4M$

The freezing point depends on the Van't Hoff factor.That means higher the Van't Hoff factor, higher will be the freezing point and vice-versa.

Hence, the correct option is, (b) 0.1 M $K_3[Fe(CN)_6]$