The freezing point of a $0.05 \ molal$ solution of a non-electrolyte in water is $(K_f = 1.86 \ K \ kg \ mol^{-1})$:

$CaCl_2$ is preferred over $NaCl$ for clearing ice on roads,particularly in very cold countries. This is because:

Calculate the freezing point of a $0.05 \ m$ aqueous solution of a non-electrolyte. (in $K$)

When $4.5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water,the freezing point of the solution is lowered by $0.465^o C$. The molar mass of the solute is ....... $g/mol$. (Given $K_f = 1.86 \ K \ kg \ mol^{-1}$)

$1\%$ solution of $Ca(NO_3)_2$ has a freezing point:

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 $180.75 \, J/g$.