$A$ 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:

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 a solution prepared by adding $6.1 \ g$ of benzoic acid $(C_6H_5COOH)$ in $500 \ g$ of benzene freezes at $-0.290 \ ^oC$,find the percentage of association. (Given: $K_f$ of benzene $= 5.12 \ K \ kg/mol$,freezing point of pure benzene $= 5.50 \ ^oC$,molar mass of benzoic acid $= 122 \ g/mol$) (in $\%$)

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 ........

$A$ 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 $= 100^{\circ} C$]

When $0.0106 \text{ mole}$ of acetic acid is dissolved in $1 \text{ kg}$ of water, the observed freezing point depression is $0.0205 \text{ K}$. If the calculated freezing point depression is $0.0197 \text{ K}$, the Van't Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ of acetic acid are respectively: