During the depression in freezing point experiment,an equilibrium is established between the molecules of

Calculate the molar mass of the solute when $1.5 \ g$ of a non-volatile solute is dissolved in $100 \ mL$ of a solvent having a density of $0.8 \ g \ mL^{-1}$,which lowers its freezing point by $0.75 \ K$. (Freezing point depression constant for the solvent is $5 \ K \ kg \ mol^{-1}$).

The molal depression constant for a liquid is $2.77^{\circ} C \ kg \ mol^{-1}$. What is its value on the Kelvin scale?

$A$ fixed amount of glucose is dissolved in $100 \text{ g}$ water to form a solution that freezes at $-0.2^\circ\text{C}$. If the solution is cooled down to $-0.25^\circ\text{C}$, then ......... $\text{g}$ of ice would have separated.

$1.8 \ g$ 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 $= 1.86 \ K \ kg \ mol^{-1}$)

Which among the following equations represents the relation between the cryoscopic constant,depression in freezing point,and molality?