Question

100 g of liquid A (molar mass 140 g mol⁻¹) was dissolved in 1000 g of liquid B (molar mass 180 g mol⁻¹). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Answer 1

Number of moles of liquid A, n_A = `100/140` mol

= 0.714 mol

Number of moles of liquid B, n_B = `1000/180` mol

= 5.556 mol

Then, the mole fraction of A, χ_A = `n_A/(n_A + n_B)`

= `0.714/(0.714 + 5.556)`

= 0.114

And, mole fraction of B, χ_B = 1 − 0.114 = 0.886

Vapour pressure of pure liquid B, `p_B^0` = 500 torr

Therefore, the vapour pressure of liquid B in the solution,

p_B = `p_B^0 chi_B`

= 500 × 0.886

= 443 torr

Total vapour pressure of the solution, p_total = 475 torr

∴ Vapour pressure of liquid A in the solution,

p_A = p_total − p_B

= 475 − 443

= 32 torr

Now,

p_A = `p_A^0 chi_A`

`=> p_A^0 = p_A/chi_A`

= `32/0.114`

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.

Answer 2

Number of moles of liquid A = `100/140`

= 0.714

Number of moles of liquid B = `1000/180`

= 5.555

Hence, mole fraction of A in the solution

`chi_A = 0.714/(0.714 + 5.555)`

= 0.114

and mole fraction of B in the solution

`chi_B = 5.555/(0.714 + 5.555)`

= 0.886

According to Raoult’s law,

`p_A = chi_Ap_A^circ`

`p_A = p_A^circ`    ...(i)

`p_A = chi_Bp_B^circ`

= 0.886 × 500

= 443 torr    ...(ii)

∴ Total vapour pressure of the solution

p = p_A + p_B

= `0.114 p_A^circ + 443`

Given that p = 475 torr

∴ `0.114 p_A^circ + 443 = 475`

or `p_A^circ = (475 - 443)/0.114`

= 280.7 torr

Putting this value in eq. (i), we have

p_A = 0.114 × 280.7

= 31.9998

= 32 torr