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Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 ∘C, the vapor pressure of pure benzene is 100.84 Torr. What is the vapor pressure of a solution made from dissolving 11.5 g of biphenyl in 31.9 g of benzene?

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jufsanabriasa

Answer:

[tex]P_{solution} = 85.3Torr[/tex]

Explanation:

Raoult's law is a tool that allows to determine vapour pressure of solutions. The formula is:

[tex]P_{solution} = X_{solvent}P_{solvent}[/tex] (1)

Where

P is Pressure of solution and solvent and X is mole fraction.

Moles of solute and solvent are:

Biphenyl:

11.5g×(1mol /154.21g) = 0.0746mol

Benzene

31.9g×(1mol /78.11g) = 0.408mol

Mole fraction of benzene is:

[tex]\frac{0.408mol}{0.408mol + 0.0746mol}[/tex] = 0.846

Replacing in (1):

[tex]P_{solution} = 0.846*100.84Torr[/tex]

[tex]P_{solution} = 85.3Torr[/tex]

I hope it helps!

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