The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is

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52.3%

60.23%

83.5%

46.7%

answer is B.

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Detailed Solution

$i = \frac{C a l c u l a t e d v a l u e o f M o l e c u l a r w t}{o b s e r v e d v a l u e o f M o l e c u l a r w t}$
$i = \frac{170}{92.64} = 1.835 α = \frac{i - 1}{n - 1} \times 100 = \frac{1.835 - 1}{2 - 1} \times 100 = 83.5$

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