SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.
First, we will calculate the partial pressure of SF₆ using the following expression.
[tex]pSF_6 = P \times \frac{ppb}{10^{9} }[/tex]
where,
[tex]pSF_6 = P \times \frac{ppb}{10^{9} } = 1 atm \times \frac{1.0 ppb}{10^{9} } = 1.0 \times 10^{-9} atm[/tex]
Then, we will convert 1.0 cm³ to L using the following conversion factors:
[tex]1.0 cm^{3} \times \frac{1mL}{1cm^{3}} \times \frac{1L}{1000 mL} = 1.0 \times 10^{-3} L[/tex]
Next, we will convert 46 °C to Kelvin using the following expression.
[tex]K = \° C + 273.15 = 46 + 273.15 = 319 K[/tex]
Afterward, we calculate the moles (n) of sulfur hexafluoride using the ideal gas equation.
[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1.0 \times 10^{-9} atm \times 1.0 \times 10^{-3} L}{(0.082 atm.L/mol.K) \times 319 K} = 3.8 \times 10^{-14} mol[/tex]
Finally, we will convert 3.8 × 10⁻¹⁴ mol to molecules using Avogadro's number.
[tex]3.8 \times 10^{-14} mol \times \frac{6.02 \times 10^{23} molecules }{mol} = 2.3 \times 10^{10} molecules[/tex]
SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.
You can learn more about partial pressure here: https://brainly.com/question/13199169
