The cryoscopic constant of a liquid is the ratio of the depression of freezing point and the........

If a $0.4 \text{ molal}$ solution of a nonvolatile solute in an organic solvent decreases its freezing point by $2.4 \text{ K}$, then the molal depression constant $(K_f)$ of the solvent will be:

$0.05 \ mol$ of a non-volatile solute is dissolved in $500 \ g$ of water. What is the depression in freezing point of the resultant solution (in $K$)? $(K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1})$

Calculate the molality of a solution having freezing point depression $3.6 \ K$ and freezing point depression constant $4.8 \ K \ kg \ mol^{-1}$.

$A$ solvent freezes at $17^{\circ} C$ and it has a latent heat of fusion of $180 \ J \ g^{-1}$. The molal depression constant $(K_{f})$ of the solvent is (in $K \ kg \ mol^{-1}$):

The normal freezing point $(f.p.)$ of nitrobenzene is $278.82 \ K$. If a non-volatile solute is dissolved in it and a solution of $0.25 \ molality$ is prepared,the observed freezing point of the solution is $276.82 \ K$. Evaluate the cryoscopic constant $(K_f)$ of nitrobenzene in $K \ kg \ mol^{-1}$.