The van't Hoff factor $i$ for a compound which undergoes dissociation in one solvent and association in another solvent is respectively:

The degree of dissociation $(\alpha)$ of a weak electrolyte,$A_xB_y$ is related to van't Hoff factor $(i)$ by the expression

$2$ molal solution of a weak acid $HA$ has a freezing point of $-3.885^{\circ} C$. The degree of dissociation of this acid is ........ $\times 10^{-3}$. (Round off to the Nearest Integer).
[Given: Molal depression constant of water = $1.85 \ K \ kg \ mol^{-1}$,Freezing point of pure water = $0^{\circ} C$]

$1.7920 \ g$ of $K_2SO_4$ is present in $1000 \ mL$ of solution. If such solution has an osmotic pressure of $0.680 \ bar$ at $26 \ ^\circ C$,then find out the van't Hoff factor value.

Pure benzene freezes at $5.3^oC$. $A$ solution of $0.223 \ g$ of $C_6H_5CH_2COOH$ in $4.49 \ g$ of benzene freezes at $4.47^oC$. Given $K_f = 5.12 \ K \ kg/mol$,what can be concluded from this?

If $\alpha$ is the degree of dissociation of $Na_2SO_4$,the Van't Hoff factor $(i)$ used for calculating the molecular mass is