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

Benzoic acid undergoes dimerization in benzene. $x \ g$ of benzoic acid (molar mass $122 \ g \ mol^{-1}$) is dissolved in $49 \ g$ of benzene. The depression in freezing point is $1.12 \ K$. If the degree of association of the acid is $88 \%$,what is the value of $x$? ($K_f$ for benzene $= 4.9 \ K \ kg \ mol^{-1}$)

$A$ solute $A$ dimerizes in water. The boiling point of a $2 \, M$ solution of $A$ is $100.52^{\circ} C$. The percentage association of $A$ is ..... $\%$.
(Round off to the Nearest integer)
[Use : $K_{b}$ for water $= 0.52 \, K \, kg \, mol^{-1}$,Boiling point of water $= 100^{\circ} C$]

Pure benzene freezes at $5.3\,^oC$. $A$ solution of $0.223\, g$ of phenylacetic acid $(C_6H_5CH_2COOH)$ in $4.4\, g$ of benzene $(K_f = 5.12\, K\, kg\, mol^{-1})$ freezes at $4.47\,^oC$. From this,it can be concluded that phenylacetic acid ........

$75.2 \ g$ of phenol is added to $1 \ kg$ of solvent. The depression in freezing point is $7 \ K$. If phenol undergoes dimerization,calculate the percentage of association. $(K_f = 14 \ K \ kg \ mol^{-1})$

At $300 \ K$, the osmotic pressure of a decinormal solution of sodium chloride is $4.82 \ atm$. The degree of dissociation of sodium chloride is $x \times 10^{-2}$. The value of $x$ is $\left(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}\right)$