#### 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.

#### Solution

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, mole fraction of A, `x_A = n_A/(n_A+n_B)`

`= 0.714/(0.714+5.556)`

= 0.114

And, mole fraction of B, *x*_{B} = 1 − 0.114

= 0.886

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

Therefore, vapour pressure of liquid B in the solution,

`p_B= p_B^0x_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^0x_A`

`=>p_A^0 = p_A/x_A`

`= 32/0.114`

= 280.7 torr

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