Question

20.A 0.2 molal solution of KCl freezes at -0.68°C. If Kf ( depression in freezing point) for H.O( it's H2o)is 1.86, the
degree of dissociation of KCl is
1) 85%
2) 65%
3) 83%
4) 90%
who will answer it correctly will be awarded 40 points and I will mark as brainliest
the answer should be with explanation
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Answer 1

Answer: 3) 83%

Explanation:

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

Freezing point of the solution,$T_f$ = -0.68°C

Freezing point of water = $T_f^0$  = 0°C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

$\Delta T_f=T-T_f=0^oC-(-0.68^oC)=0.68^oC$

m = molality = 0.2 m

i = vant hoff factor = ?

$0.68^oC=i\times 1.86 ^oC/(mol/kg)\times 0.2$

$i=1.83$

$i=\frac{\text {observed colligative property}}{\text {Calculated Colligative property}}$

$KCl\rightarrow K^{+}++Cl^{-}$

0.2          0             0

$0.2(1-\alpha)$     $0.2\alpha$  $0.2\alpha$  

Total moles after dissociation =$0.2+0.2\alpha$  

thus $i=\frac{0.2+0.2\alpha}{0.2}$

$1.83=\frac{0.2+0.2\alpha}{0.2}$

$\alpha=0.83$

Thus percent dissociation is 83%