Question

question 7
a mixture of three noble gases (he, ne and ar) is confined to a 4.683 l
container at 25 °c. the he exerts a pressure 2.97 atm, the ne exerts a
pressure of 1.18 atm and the ar exerts a pressure of 3.39 atm.
what is the total pressure in the container? atm
what is the mole fraction of ne?
what is the mass of ar in the gas mixture? g

Explanation:

Part 1: Calculate the total pressure

Step1: Apply Dalton's law of partial pressures

Dalton's law states that \(P_{total}=P_{He}+P_{Ne}+P_{Ar}\).

$$P_{total}=2.97 + 1.18+3.39$$

$$P_{total}=7.54\space atm$$

Part 2: Calculate the mole fraction of Ne

Step1: Use the formula for mole fraction

The mole fraction formula is \(x_{i}=\frac{P_{i}}{P_{total}}\). For Ne, \(x_{Ne}=\frac{P_{Ne}}{P_{total}}\)

$$x_{Ne}=\frac{1.18}{7.54}$$

$$x_{Ne}\approx0.156$$

Part 3: Calculate the mass of Ar

Step1: Use the ideal gas law to find moles of Ar

The ideal gas law is \(PV = nRT\). First, convert \(T = 25^{\circ}C=(25 + 273.15)K=298.15K\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(P_{Ar}=3.39\space atm\), \(V = 4.683\space L\)

$$n_{Ar}=\frac{P_{Ar}V}{RT}=\frac{3.39\times4.683}{0.0821\times298.15}$$

$$n_{Ar}=\frac{15.87537}{24.488715}$$

$$n_{Ar}\approx0.648\space mol$$

Step2: Convert moles of Ar to mass

The molar mass of Ar is \(M = 39.948\space g/mol\). Use \(m=n\times M\)

$$m_{Ar}=0.648\times39.948$$

$$m_{Ar}\approx25.9\space g$$

Answer:

• Total pressure: \(7.54\space atm\)
• Mole fraction of Ne: \(0.156\)
• Mass of Ar: \(25.9\space g\)