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

Find the depression in freezing point of a solution when $3.2 \ g$ of a non-volatile solute with a molar mass of $128 \ g \ mol^{-1}$ is dissolved in $80 \ g$ of solvent,given that the cryoscopic constant of the solvent is $4.8 \ K \ kg \ mol^{-1}$. (in $K$)

The freezing point of a $5\%$ (by mass) solution of cane sugar in water is $271 \, K$. If the freezing point of pure water is $273.15 \, K$,the freezing point of a $5\%$ (by mass) solution of glucose in water will be .......... $K$.

$5 \ g$ sucrose (molar mass $= 342 \ g \ mol^{-1}$) is dissolved in $100 \ g$ of solvent,decreasing the freezing point by $2.15 \ K$. What is the cryoscopic constant $(K_{f})$ of the solvent?

Calculate the molar mass of a non-volatile solute when $1 \ g$ of it is dissolved in $100 \ g$ of solvent,which decreases its freezing point by $0.2 \ K$. Given: $K_{f} = 1.2 \ K \ kg \ mol^{-1}$.

At $T$ $(K)$,$x \ g$ of a non-volatile solid (molar mass $78 \ g \ mol^{-1}$) when added to $0.5 \ kg$ water,lowered its freezing point by $1.0^{\circ} C$. What is $x$ (in $g$)? ($K_{f}$ of water at $T$ $(K)$ = $1.86 \ K \ kg \ mol^{-1}$)