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

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 $H_2O = 1.86 \ K \ kg \ mol^{-1}$)

$6 \ g$ of a non-volatile,non-electrolyte $X$ dissolved in $100 \ g$ of water freezes at $-0.93^{\circ} C$. The molar mass of $X$ in $g \ mol^{-1}$ is ($K_f$ of $H_2O = 1.86 \ K \ kg \ mol^{-1}$)

If the $K_f$ value of $H_2O$ is $1.86 \ K \ kg \ mol^{-1}$,the value of $\Delta T_f$ for a $0.1 \ m$ solution of a non-volatile solute is:

$A$ solution containing $8.0 \, g$ of nicotine in $92 \, g$ of water freezes $0.925 \, ^{\circ}C$ below the normal freezing point of water. If the molal freezing point depression constant,$k_f = 1.85 \, ^{\circ}C \, kg \, mol^{-1}$,then the molar mass of nicotine is $...$

Calculate the molality of the solution of a nonvolatile solute if it freezes at $-0.36 \ ^{\circ}C$. [Given: $K_{f}$ for solvent $= 1.86 \ K \ kg \ mol^{-1}$]