Question

A quantity of 5.0 g of a mixture of He and another gas occupies a volume of 2.4 L at 300 K and 760 mm Hg. The gas freezes at 270 K. At 15 K, the pressure of the gas mixture is 19 mm Hg (at the same volume). What is the molecular mass of the gas?

• A. 96
• B. 4.0
• C. 48
• D. 192

Answer 1

Answer: (A)

96

Answer 2

The key insight is that at 15 K, only Helium contributes to the pressure because the other gas freezes.

1. Moles of Helium ($n_{He}$): At 15 K, $P_{He} = 19 \text{ mm Hg}$, $V = 2.4 \text{ L}$, $T = 15 \text{ K}$. Using $PV = nRT$ with $R = 62.36 \text{ L mm Hg K}^{-1} \text{ mol}^{-1}$: $n_{He} = \frac{19 \times 2.4}{62.36 \times 15} \approx 0.04875 \text{ mol}$

2. Mass of Helium ($m_{He}$): $m_{He} = n_{He} \times M_{He} = 0.04875 \text{ mol} \times 4.0 \text{ g/mol} = 0.195 \text{ g}$

3. Mass of Unknown Gas ($m_{unknown}$): $m_{unknown} = 5.0 \text{ g} - m_{He} = 5.0 \text{ g} - 0.195 \text{ g} = 4.805 \text{ g}$

4. Total Moles ($n_{total}$) at 300 K: At 300 K, $P_{total} = 760 \text{ mm Hg}$, $V = 2.4 \text{ L}$, $T = 300 \text{ K}$. $n_{total} = \frac{760 \times 2.4}{62.36 \times 300} \approx 0.0975 \text{ mol}$

5. Moles of Unknown Gas ($n_{unknown}$): $n_{unknown} = n_{total} - n_{He} = 0.0975 \text{ mol} - 0.04875 \text{ mol} = 0.04875 \text{ mol}$

6. Molecular Mass of Unknown Gas ($M_{unknown}$): $M_{unknown} = \frac{m_{unknown}}{n_{unknown}} = \frac{4.805 \text{ g}}{0.04875 \text{ mol}} \approx 98.56 \text{ g/mol}$

   The closest option is 96.