Respuesta :
The mole fraction of each component in the mixture is
a. He = 0.98
b. [tex]O_{2}[/tex] = 0.02
As per the given question,
An 11.0 L Scuba diving tank contains a Helium-Oxygen(heliox) mixture made up of He and [tex]O_{2}[/tex] at a temperature of 298 K.
- He = 23.5 g
- [tex]O_{2}[/tex] = 4.45 g
⇒ Number of moles of He ([tex]n_{He}[/tex]) = mass / molar mass
= 23.5 / 4 g mole
= (23.5/4)(g × mole/g)
[tex]n_{He}[/tex] = 5.875 mole
⇒ Number of moles of O2 ([tex]n_{O_{2} }[/tex]) = mass / molar mass
= 4.45 / 32 g mole
= (4.45/32)(g × mole/g)
[tex]n_{O_{2} }[/tex] = 0.1390625 mole
⇒ Mole fraction of He ([tex]X_{He}[/tex]) = [tex]n_{He} / n_{He}+ n_{O_{2} }[/tex]
= 5.875 / 5.875 + 0.1390625
= 5.875 / 6.0140625
[tex]X_{He}[/tex] = 0.97687711 ≅ 0.98
⇒ Mole fraction of O2 ([tex]X_{O_{2} }[/tex]) = [tex]n_{O_{2} } / n_{O_{2} } + n_{He}[/tex]
= 0.1390625 / 0.1390625 + 5.875
= 0.1390625 / 6.0140625
[tex]X_{O_{2} }[/tex] = 0.023122889 ≅ 0.02
Therefore, the mole fraction of He is 0.98 and the mole fraction of [tex]O_{2}[/tex] is 0.02.
To know more about finding the mole fraction of He and [tex]O_{2}[/tex] refer to:
https://brainly.com/question/29428896
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