$CaCl_2$ is preferred over $NaCl$ for clearing ice on roads,particularly in very cold countries. This is because:

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:

Find the freezing point depression of a solution having molality $0.25 \ mol \ kg^{-1}$. $(K_{f} = 4.0 \ K \ kg \ mol^{-1})$ (in $K$)

During the depression of freezing point in a solution,which of the following are in equilibrium?

If $6 \ g$ of solute dissolved in $100 \ g$ of water lowers the freezing point by $0.93 \ K$. What is the molar mass of the solute? $(K_{f} = 1.86 \ K \ kg \ mol^{-1})$

The molal depression constant for water is $1.86\,^{\circ}C/m$. If $342\,g$ of sugar $(C_{12}H_{22}O_{11})$ is dissolved in $1000\,g$ of water,the freezing point of the solution will be ............. $^{\circ}C$.