Question

A solution is made by dissolving 0.0803 g of biphenyl (C₁₂H₁₀) in 350.0 mL of benzene (C₆H₆)

A. If the boiling point of pure benzene is 80.1 °C, then what would be the boiling point of this solution? (Kb for benzene is 2.53 °C/m and the density of benzene is 0.877 g/mL)

B. 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 and the density of benzene is 0.877 g/mL)

Answer 1

Number of moles of biphenyl=amount of biphenyl/ mw of biphenyl= 0.0803g/154.21 g/mole = 0.000521 mole

Amount of benzene= volume of benzene X density of benzene= 350mL X 0.877 g/mL= 306.95 g= 0.30695 kg

Therefore molality of biphenyl solution = no of moles of biphenyl/amount of benzene in kg= 0.000521 mole/0.30695 kg = 0.001697 m

Elevation in boiling point of a solution = Kb X m

Where Kb= molal boiling point elevation constant for benzene

m= molality of solution

Hence

Elevation in boiling point of biphenyl solution = 2.53 centigrade/m X 0.001697 m= 0.004 centigrade

Therefore boiling point of biphenyl solution = boiling point of benzene + elevation in boiling point of the biphenyl solution= 80.1 centigrade + 0.004 centigrade = 80.104 centigrade

B)

Number of moles of benzene = amount of benzene/ mw of benzene = 306.95 g/78.11 g/mole = 3.9297 mole

Mole fraction of benzene = no of moles of benzene/(no of moles of benzene + no of moles of biphenyl)= 3.9297 mole ( 3.9297 mole + 0.000521 mole ) = 0.9999

We know that

Vapor pressure of biphenyl solution in benzene = mole fraction of benzene X vapor pressure of benzene

                                                                                        = 0.9999 X 24.4 kPa= 24.398 kPa