$1.00 \ g$ of a non-electrolyte solute dissolved in $50 \ g$ of benzene lowered the freezing point of benzene by $0.40 \ K$. The ${K_f}$ for benzene is $5.12 \ K \ kg \ mol^{-1}$. The molecular mass of the solute will be ............ $g \ mol^{-1}$.

$31 \ g$ of ethylene glycol $(C_2H_6O_2)$ is dissolved in $600 \ g$ of water. The freezing point depression of the solution is ($K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$) (in $K$)

The freezing point of $1 \%$ solution of lead nitrate in water will be

$83 \ g$ of ethylene glycol is dissolved in $625 \ g$ of water. The freezing point of the solution is $...... \ K$. (Nearest integer) [Use: Molal freezing point depression constant of water $= 1.86 \ K \ kg \ mol^{-1}$,Freezing point of water $= 273 \ K$,Atomic masses: $C = 12.0 \ u, O = 16.0 \ u, H = 1.0 \ u$]

$1.95 \ g$ of non-volatile and non-electrolyte solute dissolved in $100 \ g$ of benzene lowered the freezing point of it by $0.64 \ K$. The molar mass of the solute (in $g \ mol^{-1}$) is: $(K_{f}(C_6H_6) = 5.12 \ K \ kg \ mol^{-1})$

When a solute is dissolved in a solvent,the freezing point decreases by $0.184 \ ^oC$. What will be the molality of the solution? (Given $K_f = 18.4 \ K \ kg \ mol^{-1}$)