Question

0.01 molar aqueous solution of K3(FeCN)6 freeze at -0.062°c.Calculate the percentage dissociation of solute

Answer 1

Assume Kf of water=1.86

We know,

Change in bpt=i*Kf*m

=>0.062=1.86*i*0.01

=>620=186*i

=>i=20/6=10/3

We also know,

i=(a-1)/(n-1)

Use this and get the answer(here n=3+1=4 s the salt dissociates as 3K+ + [Fe(CN)6]3-)

Answer 2

Answer: 77%

Explanation:

Depression in freezing point is given by:

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

$T_f$ = change in freezing point

$k_f$ = freezing point constant

m = molality

i = Van'T Hoff factor

$T_f^0-T_f=i\times k_f\times m$

$0-(-0.062)^0C=i\times 1.86\times 0.01$

$i=3.33$

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

$\alpha=\frac{i-1}{n-1}=\frac{i-1}{4-1}$

$\alpha=\frac{i-1}{n-1}=\frac{3.33-1}{4-1}=0.77=77\%$