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

Question

The molecular weight of NaCl , determined by studying freezing point depression of its 0.5 % aqueous solution is 30 . The apparent degree of dissociation of NaCl is:- (A) 0.95 (B) 0.50 (C) 0.60 (D) 0.30

Tardigrade Admin · Accepted Answer

i=(Theoretical molar wt/Observed molar wt)=(58 . 5/30)=1.95 i=1-α +nÎ± 1.95=1-α +2α α =0.95

Q. The molecular weight of $N a Cl$ , determined by studying freezing point depression of its $0.5%$ aqueous solution is $30$ . The apparent degree of dissociation of $N a Cl$ is:-

$0.95$

$0.50$

$0.60$

$0.30$

$i = \frac{T h eore t i c a l m o l a r wt}{O b ser v e d m o l a r wt} = \frac{58.5}{30} = 1.95$
$i = 1 - α + n α$
$1.95 = 1 - α + 2 α$
$α = 0.95$

Q. The molecular weight of NaCl , determined by studying freezing point depression of its 0.5% aqueous solution is 30 . The apparent degree of dissociation of NaCl is:-

0.95

0.50

0.60

0.30