Calculate the cryoscopic constant $(K_f)$ of a solvent when $2.5 \ g$ of a solute is dissolved in $35 \ g$ of the solvent,which lowers its freezing point by $3 \ K$. (Molar mass of the solute is $117 \ g \ mol^{-1}$)

What is the molar mass (in $g \ mol^{-1}$) of a substance,which forms a $7 \%$ by mass solution in water,which freezes at $-0.93^{\circ} C$? ($K_{f}$ of $H_2O = 1.86 \ K \ kg \ mol^{-1}$)

$1.8 \ g$ 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 $^\circ C$)

The cryoscopic constant of a liquid is the ratio of the depression of freezing point and the........

$A$ $5 \%$ solution (by mass) of cane sugar in water has a freezing point of $271 \, K$. Calculate the freezing point of a $5 \%$ glucose solution in water,given that the freezing point of pure water is $273.15 \, K$.

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 $1.86 \ K \ kg \ mol^{-1}$]