Question

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 14.6 g of biphenyl in 34.2 g of benzene?

Answer 1

This problem is based on Colligative Property " Lowering of Vapor Pressure of Liquid"

Mathematically we have,

(P^o - P) / P^o = n/N --------------- (1)

where,

Solvent : Benzene and Non volatile, nonionizing solute = Biphenyl.

Vapor Pressure of Benzene P^o = 100.84 torr,

Vapor pressure of solution (Benzene+Biphenyl) = P = ?

# of moles of Solute Biphenyl = Mass of Biphenyl dissolved / Molar mass of Biphenyl = 14.6/154.21 = 0.095

# of Moles of Solvent (Benzene) = Mass of Benzene / Molar mass of Benzene = 34.2/78.11 = 0.438

Using these values in eq.(1) we get,

(100.84-P) / P = 0.095/0.438

(100.84-P) / P = 0.22

(100.84-P) = 0.22P

1.22P = 100.84

P = 100.84 / 1.22

P = 82.66 torr.

Vapor pressure of Benzene+biphenyl solution = 82.66 torr.

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