$20555 mg$ of a $1 : 1$ electrolytes (mol. wt. = 100) are dissolved in $500 g$ of water. The freezing point of the solution is $- 0.74^{o} C$ , $K_{f} = 1.8 {K m}^{- 1}$ . The degree of ionisation of electrolyte is:

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Solution

We know,
$Δ T_{f} = K_{f} \times m Δ T_{f} = \frac{1000 \times K_{f} \times w}{M \times W}$
Whre, $w = weight of solute$
$M = Molar mass of solute$
$W = weight of solvent$
$∴ 0.74 = \frac{1000 \times 1.8 \times 20.555}{M \times 500} \Rightarrow M \approx 100$
Now, $\frac{Normal mol. wt.}{exp. mol. wt.} = 1 + α = \frac{100}{100}$

$∴ α = 0$

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