Question

0.5 molal aqueous solution of a weak acid h x is 20% ionized. If k f for water is 1.86 k k g m o l 1 , the lowering in freezing point of the solution is

Answer 1

Answer : 1.12 K

Explanation:

Depression in freezing point :  $\Delta T_f=i\times k_f\times m$

where,  $\Delta T_f$ =depression in freezing point

i = vant hoff factor

$k_f$ = freezing point constant

m = molality

i =$\frac{\text {Total no on moles at equilibrium}}{\text {Initial moles}}$

$HX\rightarrow H^++X^-$

 I:      0.5                    0        0

C:    c-cα                    cα        cα

α =$\frac{20}{100}=0.2$, c = 0.5M

c-cα = 0.4 M , cα = 0.1 M, cα = 0.1 M

Total moles at equilibrium= 0.4M + 0.1M + 0.1M = 0.6M

$i=\frac{0.6M}{0.5M}=1.2$

Depression in freezing point :  $\Delta T_f=1.2\times 1.86K/kgmol\times 0.5M=1.12K$

                       

Answer 2

Explanation:

plzz refer the attachment