Question

At a certain temperature the vapor pressure of pure heptane (C_7 H_16) is measured to be 170. torr. Suppose a solution is prepared by mixing 86.7 g of heptane and 125. g of acetyl bromide (CH_3 COBr). Calculate the partial pressure of heptane vapor above this solution. Round your answer to 3 significant digits. Note for advanced students: you may assume the solution is ideal.

Answer 1

Explanation:

The given data is as follows.

       Pressure (P) = 170 torr,      mass of heptane (m) = 86.7 g

First, we will calculate the number of moles as follows.

          No. of moles = $\frac{mass}{\text{molar mass}}$

                        = $\frac{86.7 g}{100 g/mol}$

                        = 0.867 mol

Now, the number of moles of $CH_{3}COBr$ are calculated as follows.

     No. of moles =  $\frac{mass}{\text{molar mass}}$

                           =  $\frac{125}{122.9}$    

                           = 1.07

Therefore, mole fraction of heptane will be calculated as follows.

     Mole fraction = $\frac{\text{moles of heptane}}{\text{total moles}}$

                           = $\frac{0.867}{0.867 + 1.07}$

                           = $\frac{0.867}{1.937}$

                           = 0.445

Now, we will calculate the partial pressure of heptane as follows.

            $P_{A} = x_{A}P^{o}_{A}$

                 = $170 \times 0.445$      

                 = 75.65 torr

Thus, we can conclude that the partial pressure of heptane vapor above this solution is 75.65 torr.