Question

The vapor pressure of pure water at 15 °C is 12.8 mm Hg. What is the equilibrium vapor pressure of water above a mixture of 72.0 g ethanol (CH3CH2OH, molar mass = 46.07 g/mol) and 22.0 g water?

Answer 1

The vapor pressure of water above this mixture is 5.6173 mm Hg.

Step-by-step explanation:

We need to find the vapour pressure due to water in the solution.

According to Raoult's law ,

The vapour pressure due to any one component alone is the product of it's pure vapour pressure and it's mole fraction in the solution .

i.e ,

`Y_(A)= X_(A)*(P^(0)_A)`

where ,

• 
  `Y_A` is the partial vapour pressure ( mm Hg)
• 
  `X_A` is the mole fraction
• 
  `P^(0)_A` is the pure vapour pressure = 12.8 mm Hg

∴

`X_A=(n_a)/(n_a+n_(b))\\=((22)/(18))/((22)/(18)+(72)/(46.07))\\=(1.22)/(2.785)\\=0.43885`

∴

[tex]Y_A=0.43885*12.8\\Y_A=5.6173 mm Hg[tex]

The vapor pressure of water above this mixture is 5.6173 mm Hg.