Answer:
[tex]w_{Ar}=0.814[/tex]
Explanation:
Hello,
In this case, given the temperature, volume and total pressure, we can compute the total moles by using the ideal gas equation:
[tex]PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.040atm*10.0L}{0.082\frac{atm*L}{mol*K}*318K}\\ \\n=0.41mol[/tex]
Next, using the molar fraction of argon, we compute the moles of argon:
[tex]n_{Ar}=0.41mol*0.715=0.29mol[/tex]
And the moles of methane:
[tex]n_{CH_4}=0.41mol-0.29mol=0.12mol[/tex]
Now, by using the molar masses of both argon and methane, we can compute the mass percent of argon:
[tex]w_{Ar}=\frac{m_{Ar}}{m_{Ar}+m_{CH_4}}=\frac{0.29mol*\frac{40g}{1mol} }{0.29mol*\frac{40g}{1mol}+0.12mol*\frac{16g}{1mol}} \\\\w_{Ar}=0.814[/tex]
Regards.
