Question

A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760. Torr, calculate the mole fraction of Xe in the mixture.

Answer 1

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = $720 cm^{3}$ = 0.720 L

     Temperature (T) = $25^{o}C$ = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = $\frac{PV}{RT}$

                                          = $1 \times \frac{0.720 L}{0.0821 \times 298}$

                                           = 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

               x + y = 0.0294

Also,      40x + 131y = 2.966

             x = 0.0097 mol

              y = (0.0294 - 0.0097)

                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = $\frac{y}{(x+y)}$

                               = $\frac{0.0197}{0.0294}$

                              = 0.67

Therefore, the mole fraction of Xe is 0.67.