Calculate the freezing point of a solution prepared by dissolving $1.8 \ g$ of glucose $(C_6H_{12}O_6)$ in $500 \ g$ of water. The $K_f$ value for water is $1.86 \ K \ kg \ mol^{-1}$. (in $K$)

$2.0 \ g$ 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:

Freezing point of a $4 \%$ aqueous solution of $X$ is equal to the freezing point of a $12 \%$ aqueous solution of $Y$. If the molecular weight of $X$ is $A$,then the molecular weight of $Y$ is ............. $A$.

If an aqueous solution of glucose is allowed to freeze,then crystals of which substance will be separated out first?

An aqueous solution contains $5\%$ by weight of urea and $10\%$ by weight of glucose. The freezing point of the solution is .......... $^oC$. [ $K_f$ for $H_2O$ is $1.86 \ K \ kg \ mol^{-1}$ ]

$A$ 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: $K_f = 1.86 \ K \ kg \ mol^{-1}$)