Question

an electrolyte AB is 50% ionised in aqueous solution. Calculate the freezing point of 1 molal aqueous solution

Answer 1

Answer: The freezing point of the solution is $270.21$.

Explanation:

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

$\Delta T_f$ = depression in freezing point

i= Van'T Hoff factor

$K_f$ = freezing point constant = 1.86 K kg/mol

m = molality = 1

$AB\rightarrow A^+B^-$

$\alpha =\frac{i-1}{n-1}$

$\alpha$= degree of dissociation

n = number of ions on complete dissociation

$50\%=0.5=\frac{i-1}{2-1}$

i = 1.5

$\Delta T_f=1.5\times 1.86 K kg/mol\times 1$

$\Delta T_f=T^{o}_f-T_f=273 K-T_f$

$273 K-T_f=1.5\times 1.86\times 1$

$T_f=270.21K$

The freezing point of the solution is $270.21K$.