Acetic acid dissolved in benzene shows a molecular mass of

$1.2 \ mL$ of acetic acid having density $1.06 \ g \ cm^{-3}$ is dissolved in $1 \ L$ of water. The depression in freezing point observed for this concentration of acid was $0.041^{\circ} C$. The van't Hoff factor of the acid is $(K_f \text{ of water } = 1.86 \ K \ kg \ mol^{-1})$

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

If $m$ is the normal molecular weight and $x$ is the degree of dissociation of $K_4[Fe(CN)_6]$,then the abnormal molecular weight in the solution of the salt will be $:-$

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

$A$ compound $MX_2$ has observed and normal molecular masses $65.6$ and $164$ respectively. Calculate the percentage of ionization of $MX_2$ ............ $\%$