Question

What is the vapor pressure at 23 oC of a solution of 1.20 g of naphthalene, C10H8, in 25.6 g of benzene, C6H6?. The vapor pressure of pure benzene at 23 oC is 86.0 mm Hg; the vapor pressure of naphthalene can be neglected. Calculate the vapor-pressure lowering of the solution.

Answer 1

Answer:

The vapor-pressure lowering of the solution is 2,39 mmHg

Explanation:

The Raoult's law says that the vapor pressure of a solvent in a solution is always lower than the vapor pressure of the pure solvent. The vapor pressure lowering is directly proportional to the mole fraction of the solute. In mathematical terms that means:

ΔP = $x_{solute}P^{0}_{solvent}$ (1)

Where:

ΔP is the vapor-pressure lowering of the solution

$x_{solute}$ is molar fraction of solute

And $P^{0}_{solvent}$ is the capor pressure of the pure solvent (86,0 mmHg)

The moles of naphthalene are:

$1,20g*\frac{1mol}{128,1705g}=$ 9,36x10⁻³ moles solute

And the moles of benzene are:

$25,6g*\frac{1mol}{78,11g}=$ 0,328 moles solvent

Molar fraction of solute is:

$\frac{9,36x10^{-3} moles}{0,328 + 9,36x10^{-3} moles} =$ 0,0278

Replacing in (1)

ΔP = 2,39 mm Hg

I hope it helps!