Calculate the cryoscopic constant $(K_f)$ of a solvent if the depression in freezing point of a $0.4 \ m$ solution of a non-volatile solute is $1.8 \ K$.

$40 \ g$ of glucose (Molar mass $= 180 \ g \ mol^{-1}$) is mixed with $200 \ mL$ of water. The freezing point of the solution is $..... \ K$. (Nearest integer)
[Given : $K_{f} = 1.86 \ K \ kg \ mol^{-1};$ Density of water $= 1.00 \ g \ cm^{-3};$ Freezing point of water $= 273.15 \ K$]

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

If molality of the dilute solution is doubled,the value of molal depression constant $(K_f)$ will be

The amount of urea to be dissolved in $500 \ mL$ of water $(K_f = 18.6 \ K \ kg \ mol^{-1})$ to produce a depression of $0.186 \ ^oC$ in freezing point is $....... \ g$. (Assume density of water = $1 \ g/mL$)

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: