One molal solution of a carboxylic acid in benzene shows the elevation of boiling point of $1.518 \ K$. The degree of association for dimerization of the acid in benzene is $........ \%$. ($K_b$ for benzene $= 2.53 \ K \ kg \ mol^{-1}$)

$A$ solute undergoes association in a solution to form a dimer. Which of the following values of the van't Hoff factor $i$ is possible?

$A$ 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:

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 Van't Hoff factor calculated from association data is always ... than calculated from dissociation data.

Benzoic acid molecules undergo dimerisation in benzene. $2.44 \ g$ of benzoic acid when dissolved in $30 \ g$ of benzene caused a depression in freezing point of $2 \ K$. What is the percentage of association of it? (Given $K_f(C_6H_6) = 5 \ K \ kg \ mol^{-1}$; molar mass of benzoic acid $= 122 \ g \ mol^{-1}$)