$1.8 \ g$ of fructose is added to $2 \ kg$ of water. The freezing point of the solution is $(k_f = 1.86 \ K \ kg \ mol^{-1})$ (in $^\circ C$)

Calculate the molality of a solution having freezing point depression $3.6 \ K$ and freezing point depression constant $4.8 \ K \ kg \ mol^{-1}$.

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 ............. $^{\circ}C$.

$A$ solution of $36 \ g$ of glucose $(C_6H_{12}O_6)$ in $1000 \ g$ of water is cooled to $-0.5 \ ^\circ C$. How many grams of ice would have separated from the solution? (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$)

$K_{f}$ (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 $^{\circ} C$)

$6 \ g$ of a non-volatile,non-electrolyte $X$ dissolved in $100 \ g$ of water freezes at $-0.93^{\circ} C$. The molar mass of $X$ in $g \ mol^{-1}$ is ($K_f$ of $H_2O = 1.86 \ K \ kg \ mol^{-1}$)