Question

A mixture of benzene and toluene contains 30% by mass of toluene. At 30°C, vapour pressure of pure toluene is 36.7 mm Hg and that of pure benzene is 118.2 mm Hg. Assuming that the two liquids form ideal solutions, calculate the total pressure and partial pressure of each constituent above the solution at 30°C.

Answer 1

Given: Mass percentage of toluene = 30% W/W

Vapour pressure of liquid toluene `(P_1^0)` = 36.7 mm Hg

Vapour pressure of liquid benzene `(P_2^0)` = 118.2 mm Hg

To find: Partial pressures of each constituent = ?

Total pressure = ?

Formulae: (i) P₁ = `P_1^0 x_1`

(ii) P₂ = `P_2^0 x_2`

(iii) P = P₁ + P₂

Calculation: Molar mass of toluene (C₇H₈) = (7 × 12) + (8 × 1)

= 92 g mol⁻¹

Molar mass of benzene (C₆H₆) = (6 × 12) + (6 × 1)

= 78 g mol⁻¹

Now, 30% W/W toluene means 30 g toluene in 100 g solution.

Thus, mass of benzene = 100 – 30

= 70 g

Number of moles of toluene (C₇H₈):

(n_A) = `(30 g)/(92 mol)` = 0.326 mol

Number of moles of benzene (C₆H₆): 

n_B = `70/78`

= 0.897 mol

Total number of moles (n_A + n_B) = 0.326 mol + 0.897 mol

= 1.223 mol

Mole fraction of toluene `(chi_("C"_7"H"_6)) = n_A/(n_A + n_B)`

= `0.326/1.223` 

= 0.2666

Mole fraction of benzene `(chi_("C"_6"H"_6))` = 1.0 − 0.2666

= 0.7334

Now, using formula (i),

`P_("C"_7 "H"_8) = P_("C"_7"H"_8)^0 xx chi_("C"_7"H"_8)`

= 36.7 mm Hg × 0.2666

= 9.78 mm Hg

`P_("C"_6 "H"_6) = P_("C"_6"H"_6)^0 xx chi_("C"_6"H"_6)`

= 118.2 mm Hg × 0.7334

= 86.7 mm Hg 

Now, using formula (ii),

Vapour pressure of the solution,

P = `P_("C"_7 "H"_8) + P_("C"_6 "H"_6)`

= 9.78 + 86.7

= 96.48

= 96.5 mm Hg

∴ Partial pressures of toluene and benzene are 9.78 mm Hg and 86.7 mm Hg, respectively.

∴ Total pressure above the solution is 96.5 mm Hg.