Answer to Question #156882 in Physical Chemistry for Ra

Question

The vapour pressure of an aqueous solution containing 28% by mass of liquid A( molar mass=70gmol⁻¹) is 2.4×10⁵Pa at 360K. To find out the vapour pressure of pure A if that of pure water is 0.88×10⁵Pa.

Answer 1

for binary liquid mixture vapour pressure of solution p

p = p₁⁰x₁ + p₂⁰x₂ 1 for solvent here water and 2 for solute liquid A

here p=2.4 x 10⁵ pa ; p₁⁰ = 0.88 x 10⁵ pa ;

in 100g solution A present 28g

so water present 100-28 = 72g

mol of water = mass\molar mass of water = 72g\18gmol-1 = 4 mol

mol of A = mass of A \ molar mass of A = 28g\70gmol-1 = 0.4 mol

so total mol = 4+0.4 = 4.4 mol

mole fraction of A is x₂ = 0.4\4.4 = 0.091

mole fraction of water x₁ = 4\4.4 = 0.91

so we can write p₂⁰ = (p - p₁⁰x₁)\x₂

=(2.4 x 10⁵ - 0.88 x 10⁵ x 0.91) \ 0.09 pa

=10⁵ x (2.4 - 0.80) \ 0.09 pa

=10⁵ x 1.6 \ 0.09 pa

= 17.78 x 10⁵ pa (ans)