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 freezing point of a solution prepared from $1.25 \ g$ of a non-electrolyte and $20 \ g$ of water is $271.9 \ K$. If the molar depression constant is $1.86 \ K \ kg \ mol^{-1}$,then the molar mass of the solute will be:

Pure water freezes at $273 \ K$ and $1 \ bar$. The addition of $34.5 \ g$ of ethanol to $500 \ g$ of water changes the freezing point of the solution. Use the freezing point depression constant of water as $2 \ K \ kg \ mol^{-1}$. The figures shown below represent plots of vapour pressure $(V.P.)$ versus temperature $(T)$. [molecular weight of ethanol is $46 \ g \ mol^{-1}$]. Among the following,the option representing the change in the freezing point is:

Which aqueous solution has the lowest freezing point?

Identify the concentration of the solution from the following such that the values of $\Delta T_{f}$ and $K_{f}$ are the same.

Calculate the cryoscopic constant $(K_f)$ of a solvent if the depression in freezing point of a $0.3 \text{ m}$ solution of a nonelectrolyte is $0.48 \text{ K}$.