$A$ solution containing $62 \ g$ of ethylene glycol in $250 \ g$ of water is cooled to $-10 \ ^\circ C$. If $K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$,the amount of water (in $g$) separated as ice is:

The molar freezing point constant for water is $1.86\,^{\circ}C\,kg\,mol^{-1}$. If $342\,g$ of cane sugar $(C_{12}H_{22}O_{11})$ are dissolved in $1000\,g$ of water,the solution will freeze at $............\,^{\circ}C$.

$A$ solution of $36 \ g$ of glucose $(C_6H_{12}O_6)$ in $1000 \ g$ of water is cooled to $-0.5 \ ^\circ C$. How many grams of ice would have separated from the solution? (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$)

Which of the following solutions has the highest freezing point?

$1.00 \text{ g}$ 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.

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