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

Calculate the molality of a nonvolatile solution if the solution freezes at $-0.95^{\circ}C$ $[K_f \text{ for water} = 1.86 \ K \ kg \ mol^{-1}, \text{ freezing point of water} = 0^{\circ}C]$.

When $36 \ g$ of a non-volatile,non-electrolytic solute having the empirical formula $CH_2O$ is dissolved in $1.2 \ kg$ of water,the solution freezes at $-0.93 \ ^\circ C$. The molecular formula of the solute is ($K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$)

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 $T$ $(K)$ = $1.86 \ K \ kg \ mol^{-1}$)

What is the unit of cryoscopic constant?

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 $K$)