$A$ solute $A$ undergoes association in a medium as $nA \rightleftharpoons A_n$. If the degree of association is $\alpha$,then the van't Hoff factor $i$ is given by:

$0.004 \ M \ Na_2SO_4$ is isotonic with $0.01 \ M$ glucose. The degree of dissociation of $Na_2SO_4$ is ........... $\%$.

For an aqueous solution of $FeSO_4$, the experimental molecular mass is found to be $80 \ g/mol$. What is the degree of dissociation of the salt?

The difference between the boiling point and the freezing point of a $0.2 \ m$ solution of acetic acid in benzene is $75.7 \ ^\circ C$. Calculate the value of the van't Hoff factor $i$. (For benzene,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

Calculate the percent dissociation of $0.02 \ m$ solution if its freezing point depression is $0.046 \ K$. $\left[K_{f} \text{ for water } = 1.86 \ K \ kg \ mol^{-1} ; n=2\right]$ (in $\%$)

$A$ solution of $1.25 \ g$ of '$P$' in $50 \ g$ of water lowers the freezing point by $0.3^{\circ}C$. The molar mass of '$P$' is $94 \ g \ mol^{-1}$. Given $K_{f(\text{water})} = 1.86 \ K \ kg \ mol^{-1}$,the degree of association of '$P$' in water is: