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 ........

$20 \ g$ of a binary electrolyte (molar mass = $100 \ g \ mol^{-1}$) is dissolved in $500 \ g$ of water. The freezing point of the solution is $-0.74 \ ^oC$. Given $K_f = 1.86 \ K \ kg \ mol^{-1}$,the degree of dissociation of the electrolyte is ......... $\%$.

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,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

The observed and normal masses of compound $MX_2$ are $65.6$ and $164$ respectively. The percent degree of ionisation of $MX_2$ is . . . . . . $\%.$ (Nearest integer)

$1.2 \, mL$ of acetic acid is dissolved in water to make $2.0 \, L$ of solution. The depression in freezing point observed for this strength of acid is $0.0198^{\circ} C$. The percentage of dissociation of the acid is $....$ (Nearest integer)
[Given : Density of acetic acid is $1.02 \, g \, mL^{-1}$
Molar mass of acetic acid is $60 \, g \, mol^{-1}$
$K_{f}(H_{2}O) = 1.85 \, K \, kg \, mol^{-1}$]

If $0.15 \ m$ aqueous solution of $KCl$ freezes at $-0.51^{\circ} C$,calculate the van't Hoff factor of $KCl$ (cryoscopic constant of water is $1.86 \ K \ kg \ mol^{-1}$).