$1.95 \ g$ of non-volatile and non-electrolyte solute dissolved in $100 \ g$ of benzene lowered the freezing point of it by $0.64 \ K$. The molar mass of the solute (in $g \ mol^{-1}$) is: $(K_{f}(C_6H_6) = 5.12 \ K \ kg \ mol^{-1})$

Calculate the mass of ascorbic acid ($Vitamin \ C$,$C_6H_8O_6$) to be dissolved in $75 \ g$ of acetic acid to lower its melting point by $1.5^{\circ} C$. $\left[K_f = 3 \ K \ kg \ mol^{-1}\right]$ (in $g$)

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

What is the molar mass (in $g \ mol^{-1}$) of a substance,which forms a $7 \%$ by mass solution in water,which freezes at $-0.93^{\circ} C$? ($K_{f}$ of $H_2O = 1.86 \ K \ kg \ mol^{-1}$)

When $x \times 10^{-2} \ mL$ methanol (molar mass $= 32 \ g \ mol^{-1}$; density $= 0.792 \ g \ cm^{-3}$) is added to $100 \ mL$ water (density $= 1 \ g \ cm^{-3}$),the following diagram is obtained.
$x = $ . . . . . . (nearest integer)
[Given: Molal freezing point depression constant of water at $273.15 \ K$ is $1.86 \ K \ kg \ mol^{-1}$]

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 $..... \,g \,mol^{-1}$.