The freezing point of a $1\% \, Ca(NO_3)_2$ solution is ......

$1.80 \ g$ of solute $A$ was dissolved in $62.5 \ cm^{3}$ of ethanol and the freezing point of the solution was found to be $155.1 \ K$. The molar mass of solute $A$ is $.......... \ g \ mol^{-1}$.
[Given: Freezing point of ethanol is $156.0 \ K$,Density of ethanol is $0.80 \ g \ cm^{-3}$,Freezing point depression constant of ethanol is $2.00 \ K \ kg \ mol^{-1}$]

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:

In countries nearer to the polar region,the roads are sprinkled with $CaCl_{2}$. This is

An aqueous solution containing $0.2 \ g$ of a non-volatile solute '$A$' in $21.5 \ g$ of water freezes at $272.814 \ K$. If the freezing point of water is $273.16 \ K$,the molar mass (in $g \ mol^{-1}$) of solute '$A$' is $[K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}]$

$0.440 \ g$ of a substance dissolved in $22.2 \ g$ of benzene lowered the freezing point of benzene by $0.567 \ ^oC$. Calculate the molecular mass of the substance. (Given: $K_f = 5.12 \ ^oC \ kg \ mol^{-1}$)