$1.95 \ g$ 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: $(K_{f}(C_6H_6) = 5.12 \ K \ kg \ mol^{-1})$

Mass of ethylene glycol (antifreeze) to be added to $18.6 \ kg$ of water to protect the freezing point at $-24^{\circ} C$ is . . . . . . $kg$ (Molar mass in $g \ mol^{-1}$ for ethylene glycol $= 62$,$K_{f}$ of water $= 1.86 \ K \ kg \ mol^{-1}$)

$x$ moles of $CO(NH_2)_2$ are present in $1200 \ g$ of water. If the freezing point of the solution is $-4.02 \ ^oC$,calculate the value of $x$. Given $k_f \ (H_2O) = 1.86 \ K \ kg \ mol^{-1}$.

Given that $\Delta T_f$ is the depression in freezing point of the solvent in a solution of a non-volatile solute of molality $m$,the quantity $\lim_{m \to 0} \left( \frac{\Delta T_f}{m} \right)$ is equal to:

The addition of $0.643 \,g$ of a compound to $50 \,mL$ of benzene (density $= 0.879 \,g \,mL^{-1}$) lowers the freezing point from $5.51^{\circ}C$ to $5.03^{\circ}C$. If the freezing point constant,$K_f$ for benzene is $5.12 \,K \,kg \,mol^{-1}$,the molar mass of the compound is approximately $..... \,g \,mol^{-1}$.

What is the relation between the depression in freezing point and the molar mass of a non-volatile solute?