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  4. The molecular weight of NaCl determined by studying freezing point depression of its 0.5 % aqueous solution is 30. The degree of dissociation of NaCl is .

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Q. The molecular weight of $NaCl$ determined by studying freezing point depression of its $0.5 \%$ aqueous solution is $30$. The degree of dissociation of $NaCl$ is _______.

Solutions

Solution:

$NaCl \rightarrow Na ^{+}+ Cl ^{-}\,\, (n =2$ ions $)$

$i =1+( n -1) d$

$i =1+\alpha$

Also, $i =\frac{\text { cal. mol. weight }}{\text { exp. mol. weight }}$

$=\frac{58.5}{30}=1.95$

$1+\alpha=1.95$

$\alpha=0.95$