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

$83 \ g$ 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: $C = 12.0 \ u, O = 16.0 \ u, H = 1.0 \ u$]

$1.00 \text{ g}$ of a non-electrolyte solute dissolved in $50 \text{ g}$ of benzene lowered the freezing point of benzene by $0.40 \text{ K}$. The freezing point depression constant of benzene is $5.12 \text{ K kg mol}^{-1}$. Find the molar mass of the solute.

The freezing point of an aqueous solution containing $17 \ g$ of $C_2H_5OH$ in $1000 \ g$ of water is ......... $^oC$. (Given $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$)

What is the relation between the depression in freezing point and the molar mass of a non-volatile solute?

Ethylene glycol is used as an antifreeze in cold climates. To prevent $4 \ kg$ of water from freezing at $-6^{\circ}C$,how many grams of ethylene glycol must be added? (Given for water: $K_f = 1.86 \ K \ kg \ mol^{-1}$,Molar mass of ethylene glycol $= 62 \ g \ mol^{-1}$)