If $0.072 \ g-atom$ of sulfur is dissolved in $100 \ g$ of solvent $(K_f = 7.00)$,the depression in freezing point is $0.84 \ ^\circ C$. The molecular formula of sulfur in the solution is .............

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

Calculate the molal depression constant of a solvent which has freezing point $16.6\,^oC$ and latent heat of fusion $180.75\,J\,g^{-1}$.

Calculate the molality of a solution if its depression in freezing point is $0.18 \ K$. $[K_f = 1.6 \ K \ kg \ mol^{-1}]$ (in $m$)

Bromoform has a normal freezing point of $7.734^{\circ} C$ and its $K_{f} = 14.4^{\circ} C / m$. $A$ solution of $2.60 \ g$ of an unknown substance in $100 \ g$ of bromoform freezes at $5.43^{\circ} C$. What is the molecular weight of the unknown substance?

$1.8 \ g$ of glucose (molar mass $180 \ g \ mol^{-1}$) is dissolved in $0.1 \ kg$ of water. The freezing point of the solution (in $^{\circ}C$) is ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$)