Question

A solution is made by dissolving 0.0607 g of biphenyl (C 12 H 10 ) in 350.0 mL of benzene (C 6 H 6 ) . If the vapor pressure of pure benzene is 24.4 kPa at 40.0 °C, then what will the vapor pressure of the solution be? (consider biphenyl to be nonvolatile)

Answer 1

Sol.

As Mass of biphenyl = 0.0607 g

Molar Mass of biphenyl = 154.21 g/mol

So , Moles of biphenyl

= Mass of biphenyl / Molar Mass of biphenyl  

= 0.0607 g / 154.21 g/mol  

= 0.00039 mol

Now , Volume  of benzene = 350 mL  

Density of benzene = 0.879 g/mL  

Molar Mass of benzene = 78.11 g/mol

So , Moles of benzene  

= Volume of benzene × Density of benzene / Molar Mass of benzene  

= 350 mL × 0.879 g/mL / 78.11 g/mol  

= 3.9387 mol

And Mole fraction of benzene  

= Moles of benzene / ( Moles of benzene + Moles of biphenyl )  

= 3.9387 mol / ( 3.9387 mol + 0.00039 mol )

= 0.9999

As Vapour presssure of pure benzene = 24.4 KPa  

Therefore , Vapour Pressure of solution

= Mole fraction of benzene × Vapour pressure of pure benzene

= 0.9999 × 24.4 KPa  

= 24.3975 KPa

=   24.4 KPa   ( rounded off to 3 significant figures )