Question

Pentane $\left(C_5{H}_{12}\right)$ and hexane $\left(C_6{H}_{14}\right)$combine to form an ideal solution. At25⁰Cthe vapor pressures of pentane and hexane are 511 and 150. torr, respectively. A solution is prepared by mixing 25 mL of pentane (density 50.63 g/mL) with 45 mL of hexane (density 50.66 g/mL).

a. What is the vapor pressure of this solution?

b. What is the mole fraction of pentane in the vapor that is in equilibrium with this solution?

Short answer

(a) The vapor pressure of the pentane and hexane solution will be 290.79 torr

(b) the mole fraction of pentane in the equilibrium with this solution is 1.75.

Step-by-step solution

 Step 1: Vapour pressure of the solution

In the given question, it is given that

The vapor pressure of pentane ($\text{P}{°}_{\text{Pentane}}$) = 511 torr

The vapor pressure of hexane ($\text{P}{°}_{\text{Hexane}}$) = 511 torr

The density and Volume of Pentane is 50.63 g/mL and 25 mL

Thus, the mass of pentane will be = $\text{Density}\times \text{volume=50.63}\times \text{25=1265.75g}$

The density and Volume of Pentane is 50.66 g/mL and 45 mL

Thus, the mass of pentane will be = $\text{Density}\times \text{volume=50.66}\times \text{45=2279.7g}$

Now, we can determine the number of moles of pentane and hexane i.e.

Moles of pentane = $\frac{\text{Mass}}{\text{Molarmass}}\text{=}\frac{1265.75}{72}\text{=17.5moles}$

Moles of hexane = $\frac{\text{Mass}}{\text{Molarmass}}\text{=}\frac{2279.7}{86}\text{=26.50moles}$

Therefore, total number of moles will be 17.5 + 26.50 = 44 moles

Mole fraction

Mole Fraction of pentane = $\frac{\text{Molesofpentane}}{\text{TotalMoles}}\text{=}\frac{17.5}{44}\text{=0.39}$

Mole Fraction of hexane =$\frac{\text{Molesofhexane}}{\text{TotalMoles}}\text{=}\frac{26.50}{44}\text{=0.60}$

Vapor pressure

According to the Raoult’s law, the vapor pressure of the solution will be:

$\begin{array}{l}{\text{P=P}}_1°{\text{x}}_1{\text{+P}}_2°{\text{x}}_2\\ \text{P=511}\times \text{0.39+150}\times \text{0.61=290.79torr}\end{array}$

Mole fraction of the pentane

In equilibrium with this solution of pentane and hexane,the mole fraction of pentane in the vapor will be:

$\text{x=}\frac{\text{Partialpressureofpentane}}{\text{Totalpressure}}\text{=}\frac{511}{290.79}\text{=1.75}$