The molal depression constant for a liquid is $2.77^{\circ} C \ kg \ mol^{-1}$. What is its value on the Kelvin scale?

What is the mass of solute having molar mass $60 \ g \ mol^{-1}$ when dissolved in $98 \ g$ of solvent decreases its freezing point by $0.2 \ K$ (in $g$)? (The numerical value of cryoscopic constant of solvent is $1.71 \ K \ kg \ mol^{-1}$)

$6 \ g$ 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 $H_2O = 1.86 \ K \ kg \ mol^{-1}$)

The addition of $0.643 \,g$ of a compound to $50 \,mL$ of benzene (density $= 0.879 \,g \,mL^{-1}$) lowers the freezing point from $5.51^{\circ}C$ to $5.03^{\circ}C$. If the freezing point constant,$K_f$ for benzene is $5.12 \,K \,kg \,mol^{-1}$,the molar mass of the compound is approximately $..... \,g \,mol^{-1}$.

$1.8 \ g$ of glucose (molar mass $180 \ g \ mol^{-1}$) is dissolved in $0.1 \ kg$ of water. The freezing point of the solution (in $^{\circ}C$) is ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$)

Two solutions $A$ and $B$ are prepared by dissolving $1 \ g$ of non-volatile solutes $X$ and $Y$ respectively in $1 \ kg$ of water. The ratio of depression in freezing points for $A$ and $B$ is found to be $1: 4$. The ratio of molar masses of $X$ and $Y$ is.