Answer:
[tex]M_{mix}=3.92\frac{g}{mol}[/tex]
Explanation:
Hello!
In this case, since the molar mass of a gas mixture can be computed as a weighted average containing mole fractions:
[tex]M_{mix}=M_{O_2}*x_{O_2}+M_{N_2}*x_{N_2}+M_{He}*x_{He}[/tex]
Now, since we are given the by mass percent of each gas, by assuming 1 g of mixture, we therefore have 0.20 g of oxygen, 0.2 g of nitrogen and 0.6 g of helium; next we compute the moles of each gas:
[tex]n_{O_2}=0.20g*\frac{1mol}{32g}=0.00625mol\\\\ n_{N_2}=0.20g*\frac{1mol}{28.02}=0.00714mol\\\\n_{He}=0.60g*\frac{1mol}{4.00}=0.15mol[/tex]
Next, we compute the mole fractions:
[tex]x_{O_2}=\frac{0.00625}{0.00625+0.00714+0.15}=0.038\\\\ x_{O_2}=\frac{0.00714}{0.00625+0.00714+0.15}=0.044\\\\x_{He}=\frac{0.15}{0.00625+0.00714+0.15}=0.918[/tex]
Now, we compute the molar mass of the mixture as follows:
[tex]M_{mix}=32.0\frac{g}{mol}* 0.0038+28.02\frac{g}{mol}*0.0044+4.00\frac{g}{mol}*0.918\\\\M_{mix}=3.92\frac{g}{mol}[/tex]
Best regards!
