🤓 Based on our data, we think this question is relevant for Professor Schurmeier's class at UCSD.

We’re being asked to determine the **vapor pressure of a solution**** ****contains twice as many moles of heptane as octane?**

Recall that the vapor pressure of a solution can be given by ** Raoult’s Law**.

The given solutions are composed of** heptane and octane**, which are both volatile. For a solution with volatile solute and solvent, Raoult’s Law is given as:

$\overline{){{\mathbf{P}}}_{\mathbf{solution}\mathbf{}}{\mathbf{=}}{\mathbf{}}\mathbf{\left(}\mathbf{P}{\mathbf{\xb0}}_{\mathbf{heptane}}{\mathbf{X}}_{\mathbf{heptane}}\mathbf{\right)}{\mathbf{}}{\mathbf{+}}{\mathbf{}}\mathbf{\left(}\mathbf{P}{\mathbf{\xb0}}_{\mathbf{octane}}{\mathbf{X}}_{\mathbf{octane}}\mathbf{\right)}}$

where P˚ = vapor pressure of pure component and χ = mole fraction of component. Recall that the ** mole fractions in a solution add up to 1**. Since the solution is composed of heptane and octane, this means:

$\overline{){{\mathbf{X}}}_{\mathbf{octane}\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{1}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}{{\mathbf{X}}}_{{\mathbf{hexane}}}}$

At 40°C, heptane has a vapor pressure of about 91.5 torr and octane has a vapor pressure of about 31.2 torr. Assuming ideal behavior, what is the vapor pressure of a solution that contains twice as many moles of heptane as octane?

A. 81.8 torr

B. 51.3 torr

C. 10.4 torr

D. 71.4 torr

E. 61.0 torr