$20 \ g$ of naphthoic acid $(C_{11}H_8O_2)$ dissolved in $50 \ g$ of benzene $(K_f = 1.72 \ K \ kg \ mol^{-1})$ shows a depression in freezing point of $2 \ K$. The Van't Hoff factor is?

The molecular mass of acetic acid $(CH_3COOH)$ dissolved in water is $60$,whereas when dissolved in benzene,it is $120$. This difference in behavior is because:

$0.2$ molal aqueous solution of $KCl$ freezes at $-0.680^{\circ} C$. Calculate van't Hoff factor for this solution. $(K_{f} = 1.86 \ K \ kg \ mol^{-1})$

The experimental depression in freezing point of a dilute solution is $0.025 \ K$. If the van't Hoff factor $(i)$ is $2.0$,the calculated depression in freezing point (in $K$) is

If the degree of dissociation of an aqueous solution of a weak monobasic acid is determined to be $0.3$,then the observed freezing point depression will be $.....\%$ higher than the expected/theoretical freezing point depression. (Nearest integer)

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 $\%$)