How much glucose $(molecular \ weight = 180 \ g/mol)$ should be added to $200 \ g \ H_2O$ so that when the solution is cooled to $-0.5^{\circ}C$,$14 \ g$ of ice separates out of the solution: [$K_f = 1.86 \ K \ kg/mol$ and melting point of $H_2O = 0^{\circ}C$] (in $g$)

The freezing point of $1 \%$ solution of lead nitrate in water will be

Two solutions $A$ and $B$ are prepared by dissolving $1 \ g$ of non-volatile solutes $X$ and $Y$ respectively in $1 \ kg$ of water. The ratio of depression in freezing points for $A$ and $B$ is found to be $1: 4$. The ratio of molar masses of $X$ and $Y$ is.

The addition of $0.643 \,g$ of a compound to $50 \,mL$ of benzene (density $= 0.879 \,g \,mL^{-1}$) lowers the freezing point from $5.51^{\circ}C$ to $5.03^{\circ}C$. If the freezing point constant,$K_f$ for benzene is $5.12 \,K \,kg \,mol^{-1}$,the molar mass of the compound is approximately $..... \,g \,mol^{-1}$.

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

Mass of ethylene glycol (antifreeze) to be added to $18.6 \ kg$ of water to protect the freezing point at $-24^{\circ} C$ is . . . . . . $kg$ (Molar mass in $g \ mol^{-1}$ for ethylene glycol $= 62$,$K_{f}$ of water $= 1.86 \ K \ kg \ mol^{-1}$)