Question

A very small amount of a non-volatile solute (that does not dissociate) is dissolved in 56.8 cm³ of benzene (density = 0.889 g cm⁻³). At room temperature, vapour pressure of this solution is 98.88 mm Hg while that of benzene is 100 mm Hg. Find the molality of this solution. If the freezing point of this solution is 0.73 degree lower than that of benzene, what is the value of molal freezing point depression constant of benzene?

Answer 1

Mass of benzene = Density × Volume

= 0.889 × 56.8

= 50.5 g

= 0.0505 kg

∴ Moles of benzene = `50.5/78`    ...(∵ mol. mass of benzene = 78)

= 0.647

Vapour pressure of pure benzene (p°) = 100 mm Hg

Vapour pressure of solution (p) = 98.88 mm Hg 

As the solution is dilute, we have according to the Raoult’s law,

`(p^circ - p)/p^circ = n/N`

or `(100 - 98.88)/100 = n/0.647`

`n = ((100 - 98.88)/100) xx 0.647`

= 7.25 × 10⁻³

∴ `"Molality of solution" = "No. of  moles of solute"/"Mass of solvent in kg"`

= `(7.25 xx 10^-3)/0.0505`

= 0.1436

Further, ΔT_f = K_f × m

∴ 0.73 = K_f × 0.1436

or `K_f = 0.73/0.1436`

= 5.08 K kg mol⁻¹