The freezing point of a $0.05 \ molal$ solution of a non-electrolyte in water is $(K_f = 1.86 \ K \ kg \ mol^{-1})$:

$1.00 \ g$ of a non-electrolyte solute dissolved in $50 \ g$ of benzene lowered the freezing point of benzene by $0.40 \ K$. The ${K_f}$ for benzene is $5.12 \ K \ kg \ mol^{-1}$. The molecular mass of the solute will be ............ $g \ mol^{-1}$.

If a $0.4 \text{ molal}$ solution of a nonvolatile solute in an organic solvent decreases its freezing point by $2.4 \text{ K}$, then the molal depression constant $(K_f)$ of the solvent will be:

How many grams of urea must be dissolved in $500 \ g$ of water to produce a freezing point depression of $0.186^oC$? (Given $K_f = 1.86^oC \ kg \ mol^{-1}$)

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?

$A$ camphor sample melts at $176^{\circ} C$. $K_f$ for camphor is $40 \ K \ kg \ mol^{-1}$. $A$ solution of $0.02 \ g$ of a hydrocarbon in $0.8 \ g$ of camphor melts at $156.77^{\circ} C$. The hydrocarbon is made up of $92.3 \%$ of carbon. What is the molecular formula of the hydrocarbon?