Question

Two liquids form an ideal solution. At 300 K, the vapour pressure of a solution containing 1 mol of A and 3 mol of B is 550 mm Hg. At the same temperature, if 1 mol more of B is added to this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of B in pure state. (as mm)

Answer 1

Given,
Total vapour pressure of solution = 550 mm Hg
Moles of A = 1
Moles of B = 3
Total number of moles = 1+3 = 4
Mole fraction of A = $\frac{1}{4}$
Mole fraction of B = $\frac{3}{4}$

Let the vapour pressure of A = $p_1^o$
and vapour pressure of B = $p_2^o$

Total VP=VP of A $\times$ Mole fraction of A+VP of B$\times$ Mole fraction of B ......(i)
$550=p_1^o\times \frac{1}{4}+p_2^o\times \frac{3}{4}$
$4\times 550=p_1^o+3p_2^o$ ....(ii)
On adding 1 mol of B
Moles of B becomes =3+1 =4
Moles of A =1
Total number of moles =4+1=5
Mole fraction of A =$\frac{1}{5}$
Mole fraction of B=$\frac{4}{5}$
and VP becomes =550+10=560 mm Hg
From equation (i)
$560=p_1^o\times \frac{1}{5}+p_2^o\times \frac{4}{5}$
$560\times 5=p_1^o+4p_2^o$ ...(iii)
From equations (ii) and (iii) we get,
$p_1^o$= 400 mm Hg
$p_2^o$ =600 mm Hg
Thus, vapour pressure of pure A =400 mm Hg
Vapour pressure of pure B=600 mm Hg