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

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

The freezing point of a $0.01 \ M$ aqueous glucose solution at $1 \ \text{atmosphere}$ is $-0.18 \ ^oC$. To it,an addition of an equal volume of $0.002 \ M$ glucose solution will produce a solution with a freezing point of nearly ........... $^oC$.

The freezing point of a solution prepared from $1.25 \ g$ of a non-electrolyte and $20 \ g$ of water is $271.9 \ K$. If the molar depression constant is $1.86 \ K \ kg \ mol^{-1}$,then the molar mass of the solute will be:

When $x \times 10^{-2} \ mL$ methanol (molar mass $= 32 \ g \ mol^{-1}$; density $= 0.792 \ g \ cm^{-3}$) is added to $100 \ mL$ water (density $= 1 \ g \ cm^{-3}$),the following diagram is obtained.
$x = $ . . . . . . (nearest integer)
[Given: Molal freezing point depression constant of water at $273.15 \ K$ is $1.86 \ K \ kg \ mol^{-1}$]

$40 \ g$ of glucose (Molar mass $= 180 \ g \ mol^{-1}$) is mixed with $200 \ mL$ of water. The freezing point of the solution is $..... \ K$. (Nearest integer)
[Given : $K_{f} = 1.86 \ K \ kg \ mol^{-1};$ Density of water $= 1.00 \ g \ cm^{-3};$ Freezing point of water $= 273.15 \ K$]