$0.05 \ mol$ of a non-volatile solute is dissolved in $500 \ g$ of water. What is the depression in freezing point of the resultant solution (in $K$)? $(K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1})$

During the depression in freezing point experiment, an equilibrium is established between the molecules of

$A$ solution containing $62 \ g$ of ethylene glycol in $250 \ g$ of water is cooled to $-10 \ ^\circ C$. If $K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$,the amount of water (in $g$) separated as ice is:

Lead storage battery contains $38\%$ by weight solution of $H_2SO_4$. The van't Hoff factor is $2.67$ at this concentration. The temperature in Kelvin at which the solution in the battery will freeze is $..........$ (Nearest integer). Given $K_f = 1.8 \, K \, kg \, mol^{-1}$.

Calculate the freezing point of a $0.05 \ m$ aqueous solution of a non-electrolyte. (in $K$)

$K_{f}$ (water) $= 1.86 \ K \ kg \ mol^{-1}$. The temperature at which ice begins to separate from a mixture of $10$ mass $\%$ ethylene glycol is (in $^{\circ} C$)