$0.05 \ mol$ of a non-volatile solute is dissolved in $500 \ g$ of water. What is the depression in freezing point of the resultant solution (in $K$)? $(K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1})$

Calculate the depression in freezing point of a solution when $4 \,g$ of a nonvolatile solute with a molar mass of $126 \,g \,mol^{-1}$ is dissolved in $80 \,mL$ of water. $[$Cryoscopic constant of water $K_f = 1.86 \,K \,kg \,mol^{-1}]$ (in $\,K$)

$2 \ g$ of a non-electrolyte solute (molar mass is $500 \ g \ mol^{-1}$) was dissolved in $57.3 \ g$ of xylene. If the freezing point depression constant $K_f$ of xylene is $4.3 \ K \ kg \ mol^{-1}$,then the depression in freezing point of xylene is.......... (in $K$)

Ethylene glycol is used as an anti-freezing agent. Calculate the amount of ethylene glycol to be added to $4 \ kg$ of water to prevent it from freezing at $-6 \ ^{\circ}C$. ($K_f$ for water $= 1.85 \ K \ kg \ mol^{-1}$) (molar mass of $(CH_2OH)_2 = 62 \ g \ mol^{-1}$) .......... $g$.

Calculate the molar mass of the solute when $1.5 \ g$ of a non-volatile solute is dissolved in $100 \ mL$ of a solvent having a density of $0.8 \ g \ mL^{-1}$,which lowers its freezing point by $0.75 \ K$. (Freezing point depression constant for the solvent is $5 \ K \ kg \ mol^{-1}$).

$A$ solution of sucrose (molar mass $= 342 \, g \, mol^{-1}$) has been prepared by dissolving $68.5 \, g$ of sucrose in $1000 \, g$ of water. The freezing point of the solution obtained will be ......... $^oC$. ($K_f$ for water $= 1.86 \, K \, kg \, mol^{-1}$)