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 $\%$)

When $2.44 \ g$ of benzoic acid $(C_6H_5COOH)$ is dissolved in $25 \ g$ of benzene, it shows a depression of freezing point equal to $2.2 \ K$. The molal depression constant of benzene is $5.0 \ K \ kg \ mol^{-1}$. What is the percentage association of the acid, if it forms a dimer in the solution (in $\%$)?

If the van't Hoff factor of a weak electrolyte $A_xB_y$ is $i$,then the degree of dissociation $(\alpha)$ is represented by which equation?

The observed osmotic pressure of a solution of benzoic acid in benzene is less than its expected value because:

$A$ 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:

$2 \ g$ of benzoic acid $(M_w = 122 \ g/mol)$ dissolved in $25 \ g$ of $C_6H_6$ shows a depression in freezing point equal to $1.62 \ K$. Given $K_f (C_6H_6) = 4.9 \ K \ kg \ mol^{-1}$. If the acid forms a dimer in the solution,then the percentage association of the acid is $......... \%$.