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 .

Question

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 . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

NaCl arrow Na ++ Cl - (n =2 ions ) i =1+( n -1) d i =1+α Also, i =( text cal. mol. weight / text exp. mol. weight ) =(58.5/30)=1.95 1+α=1.95 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 degree of dissociation of $N a Cl$ is _______.

$N a Cl \to N a^{+} + C l^{-} (n = 2$ ions $)$

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

$i = 1 + α$

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

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

$1 + α = 1.95$

$α = 0.95$

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 _______.

NaCl→Na++Cl−(n=2 ions ) i=1+(n−1)d i=1+α Also, i= exp. mol. weight cal. mol. weight ​ =3058.5​=1.95 1+α=1.95 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 degree of dissociation of $N a Cl$ is _______.

$N a Cl \to N a^{+} + C l^{-} (n = 2$ ions $)$

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

$i = 1 + α$

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

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

$1 + α = 1.95$

$α = 0.95$