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 $............\,^{\circ}C$.

Which of the following will have the highest freezing point $(F.P.)$ at one atmosphere?

Column-$I$ (Various solutions)	Column-$II$ (Freezing point)
$a$. $0.1 \, M \ BaCl_2$ solution	$p$. $271 \, K$
$b$. $0.1 \, M \ NaCl$ solution	$q$. $270 \, K$
$c$. $0.1 \, M \ K_3[Fe(CN)_6]$ solution	$r$. $268 \, K$
$d$. $0.1 \, M \ Al_2(SO_4)_3$ solution	$s$. $269 \, K$

Given: Freezing point of $0.1 \, M$ sucrose solution $= 272 \, K$ and freezing point of water $= 273 \, K$.
Which of the following options shows the correct matches?

Calculate the molality of the solution of a nonvolatile solute if it freezes at $-0.36 \ ^{\circ}C$. [Given: $K_{f}$ for solvent $= 1.86 \ K \ kg \ mol^{-1}$]

$A$ solution containing $8.0 \, g$ of nicotine in $92 \, g$ of water freezes $0.925 \, ^{\circ}C$ below the normal freezing point of water. If the molal freezing point depression constant,$k_f = 1.85 \, ^{\circ}C \, kg \, mol^{-1}$,then the molar mass of nicotine is $...$

$1.00 \ g$ of a non-electrolyte solute (molar mass $250 \ g \ mol^{-1}$) was dissolved in $51.2 \ g$ of benzene. If the freezing point depression constant,$K_f$ of benzene is $5.12 \ K \ kg \ mol^{-1},$ the freezing point of benzene will be lowered by .......... $K$.