$A$ solution of $5 \ g$ nonvolatile solute in $50 \ g$ water decreases its freezing point by $0.2 \ K$. Calculate the molar mass of solute if $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$.

What happens to the freezing point of benzene when a small quantity of naphthalene is added to it?

$2.7 \ kg$ of each of water and acetic acid are mixed. The freezing point of the solution will be $-x^{\circ} C$. Consider the acetic acid does not dimerise in water,nor dissociates in water. $x = . . . . . . .$ (nearest integer)
[Given : Molar mass of water $= 18 \ g \ mol^{-1}$,acetic acid $= 60 \ g \ mol^{-1}$]
$K_f \ H_2O = 1.86 \ K \ kg \ mol^{-1}$
$K_f$ acetic acid $= 3.90 \ K \ kg \ mol^{-1}$
Freezing point: $H_2O = 273 \ K$,acetic acid $= 290 \ K$

$2.0 \ g$ of a non-electrolyte dissolved in $100 \ g$ of benzene lowers the freezing point of benzene by $1.2 \ K$. The freezing point depression constant of benzene is $5.12 \ K \ kg \ mol^{-1}$. The molar mass of the solute is:

Calculate the molal depression constant of a solvent, which freezes at $15^{\circ}C$. The latent heat of fusion is $180.7 \ Jg^{-1}$.

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 $[K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}]$