Question

A solution contains naphthalene (C10H8)dissolved in hexane (C6H14) at a concentration of 12.84 % naphthalene by mass. Calculate the vapor pressure at 25 ∘C of hexane above the solution. The vapor pressure of pure hexane at 25 ∘Cis 151 torr.

Answer 1

Answer –

In this question we are given concentration of C10H8 = 12.84 % , partial pressure of hexane = 151 torr

We assume 100 g of solution

So, mass of C10H8 = 12.84 g, mass of hexane , C6H14 = 100 – 12.84 = 87.16 g

We need to calculate moles of each

Moles of C10H8 = 12.84 g / 128.17 g.mol^-1 = 0.100 moles

Moles of C6H14 = 87.16 g / 86.18 g.mol^-1 = 1.011 moles

So, total moles of solution = 1.011 +0.100 = 1.111 moles

So, mole fraction of hexane = moles of hexane / total moles of solution

Moles fraction of hexane = 1.011 / 1.111 = 0.9098

We know the Raoult’s law

Pressure of solution = Partial pressure of solvent * molar fraction

So, Pressure of solution = 151 torr * 0.9098

                                    = 137.4 torr