$0.2$ 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 observed and calculated molecular weights of calcium nitrate are $65.6$ and $164$ respectively. The degree of dissociation of calcium nitrate is .......... $\%$.

Assertion : The molecular weight of acetic acid determined by depression in freezing point method in benzene and water was found to be different.
Reason : Water is polar and benzene is nonpolar.

When $20 \ g$ of naphthoic acid $(C_{11}H_8O_2)$ is dissolved in $50 \ g$ of benzene $(K_{f} = 1.72 \ K \ kg \ mol^{-1})$,a freezing point depression of $2 \ K$ is observed. The van't Hoff factor $(i)$ is

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

The measured freezing point depression for a $0.1 \ m$ aqueous $CH_{3}COOH$ solution is $0.19^{\circ} C$. The acid dissociation constant $K_{a}$ at this concentration will be (Given, $K_{f}$ the molal cryoscopic constant $= 1.86 \ K \ kg \ mol^{-1}$)