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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?

Respuesta :

Answer:

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

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 ,

[tex]Y_{A}= X_{A}*(P^{0}_A)[/tex]

where ,

  • [tex]Y_A[/tex] is the partial vapour pressure ( mm Hg)
  • [tex]X_A[/tex] is the mole fraction
  • [tex]P^{0}_A[/tex] is the pure vapour pressure = 12.8 mm Hg

∴

[tex]X_A=\frac{n_a}{n_a+n_{b}}\\=\frac{\frac{22}{18}}{\frac{22}{18}+\frac{72}{46.07}}\\=\frac{1.22}{2.785}\\=0.43885[/tex]

∴

[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.

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