Respuesta :
Answer: The mole fractions of He in the mixture is 0.992.
The mole fraction of [tex]O_2[/tex] in the mixture is 0.008.
Explanation:
A gas mixture called heliox, 6.11% [tex]O_2[/tex] and 93.89% He gas by mass by mass.
Percentage of oxygen gas in a mixture by mass = 6.11%
In 100 gram mixture 6.11% of oxygen is present by mass :[tex]\frac{100\times 6.11}{100}=6.11 grams [/tex]
Percentage of helium gas in a mixture by mass = 93.89%
In 100 gram mixture 93.89% of helium is present by mass :[tex]\frac{100\times 93.89}{100}=93.89 grams [/tex]
Mole fraction of component 'x' in a mixture of 'x' and 'y'= [tex]=\chi _x=\frac{n_x}{n_x+n_y}[/tex]
[tex]n_x[/tex] = moles of gas 'x'=[tex]\frac{\text{mass of gas 'x'}}{\text{Molar mass of gas 'x'}}[/tex]
[tex]n_y[/tex] = moles of gas 'y'=[tex]\frac{\text{mass of gas 'y'}}{\text{Molar mass of gas 'y'}}[/tex]
Moles of oxygen gas ,when 6.11 grams of oxygen in present in a mixture:
[tex]n_{O_2}=\frac{6.11 g}{16g/mol}=0.3818 moles[/tex]
Moles of helium gas ,when 93.89 grams of helium in present in a mixture:
[tex]n_{He}=\frac{93.89 g}{2g/mol}=46.945 moles[/tex]
[tex]=\chi _{O_2}=\frac{n_{O_2}}{n_{O_2}+n_{He}}=\frac{0.3818 mol}{0.3818 mol+46.945 mol}=0.008[/tex]
[tex]=\chi _{He}=\frac{n_{He}}{n_{O_2}+n_{He}}=\frac{46.945 mol}{0.3818 mol+46.945 mol}=0.9919\approx 0.992[/tex]
The mole fractions of He in the mixture is 0.992.
The mole fraction of [tex]O_2[/tex] in the mixture is 0.008.
Mole fraction for Helium and Oxygen is 0.008 and 0.992 respectively.
Given Here,
Assuming 100g mixture of gases,
Helium is 93.89g
Oxygen is 6.11
Mole of oxygen = 0.381 moles
Mole of helium = 46.94 moles
Mole fraction formula
[tex]\rm \bold { X= \frac{Xa}{Xa+Xb} }[/tex]
Where,
Xa is mole fraction for Helium
Xb is mole fraction for Oxygen
Solving equation for Helium and oxygen
Mole fraction for Helium = 0.008
Mole fraction for Oxygen = 0.992
Hence, we can conclude that Mole fraction for Helium and Oxygen is 0.008 and 0.992 respectively.
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