Calculate the molal freezing point depression constant $(K_f)$ of a solvent which has a freezing point of $16.6 \, ^\circ C$ and a latent heat of fusion of $180.75 \, J/g$.

Two solutions $A$ and $B$ are prepared by dissolving $1 \ g$ of non-volatile solutes $X$ and $Y$ respectively in $1 \ kg$ of water. The ratio of depression in freezing points for $A$ and $B$ is found to be $1: 4$. The ratio of molar masses of $X$ and $Y$ is.

Calculate the molal depression constant of a solvent which has freezing point $16.6\,^oC$ and latent heat of fusion $180.75\,J\,g^{-1}$.

$A$ $5 \%$ solution (by mass) of cane sugar in water has a freezing point of $271 \, K$. Calculate the freezing point of a $5 \%$ glucose solution in water,given that the freezing point of pure water is $273.15 \, K$.

What is the depression of freezing point,when mole fraction of non-electrolyte solute in aqueous solution is $0.01$ (in $K$)? ($K_f$ of $H_2O = 1.86 \ K \ kg \ mol^{-1}$)

$1.8 \ g$ of glucose (molar mass $180 \ g \ mol^{-1}$) is dissolved in $0.1 \ kg$ of water. The freezing point of the solution (in $^{\circ}C$) is ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$)