For $1000 \ g$ of $1,4$-dioxane,the cryoscopic constant $K_f = 4.9 \ K \ kg \ mol^{-1}$. What will be the depression in freezing point for a $0.001 \ m$ solution prepared in dioxane?

In countries nearer to the polar region,the roads are sprinkled with $CaCl_{2}$. This is

The mass of ascorbic acid $(C_6H_8O_6)$ to be dissolved in $100 \ g$ of acetic acid to lower its freezing point by $1.5^{\circ}C$ in $g$ is: (Given: $K_f$ for acetic acid $= 3.9 \ K \ kg \ mol^{-1}$)

Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.

$83 \ g$ of ethylene glycol is dissolved in $625 \ g$ of water. The freezing point of the solution is $...... \ K$. (Nearest integer) [Use: Molal freezing point depression constant of water $= 1.86 \ K \ kg \ mol^{-1}$,Freezing point of water $= 273 \ K$,Atomic masses: $C = 12.0 \ u, O = 16.0 \ u, H = 1.0 \ u$]

$A$ 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 $= 1.86 \, K \, kg \, mol^{-1}$)