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A $0.5$ percent solution of potassium chloride was found to freeze at $-0.24^{circ} mathrm{C}$ . The percentage dissociation of potassium chloride is__________. (Nearest integer)

(Molal depression constant for water is $1.80 mathrm{~K} mathrm{~kg} mathrm{~mol}^{-1}$ and molar mass of $mathrm{KCl}$ is $74.6 mathrm{~g} mathrm{~mol}^{-1}$ )
Option: 1
98

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

Given , $\mathrm{KCl }$ soln is $\mathrm{0.5 \% }$

Assuming $\mathrm{100~g}$ of solution, the mass of $\mathrm{KCl }$ is $\mathrm{0.5\, g }$

Given,

Depression of freezing point is $\mathrm{-0.24^{\circ}C }$ .
$\mathrm{\mathrm{K}_{f} \: for\: \mathrm{KCl}=1.80 \, \mathrm{k} \: \mathrm{Kg} \; \mathrm{mol}^{-1} }$
Molar mass of $\mathrm{KCl = 74.6\, g\: mol^{-1}}$

Let us now calculate the molality of the solution

$\mathrm{molality \: of \: \mathrm{KCl}=\frac{\text { moles } of\: \mathrm{KCl}}{\text { mass } of \text { solvent }}=\frac{0.5/74 .6 }{(99.5 / 1000)}mol\: kg^{-1}}$ $\therefore\mathrm{molality \: of \: \mathrm{KCl}=0.06736 ~mol\: kg^{-1}}$

$\mathrm{\Delta T_{f}= i\, \, k_{f}\, m}$

$\mathrm{\Rightarrow 0.24= i\times 1.80\times 0.06736}$

$\mathrm{\Rightarrow i= \frac{0.24}{1.80\times 0.06736}= 1.98}$

Now, for the dissociation of $\text{KCl}$ ,

$\mathrm{\text{KCl}\longrightarrow K^+ + Cl^-}$

The van't Hoff factor(i) is given as

$\mathrm{ i= 1+\alpha }$

$\mathrm{1.98= 1+\alpha \: \Rightarrow \; \alpha = 0.98}$

Thus, the % dissociation of KCl is 98%

Hence 98 is the answer.

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