Question

A 10.0 L flask at 318 K contains a mixture of Ar and CH4 with a total pressure of 1.040 atm. If the mole fraction of Ar is 0.715, what is the mass percent of Ar?

Answer 1

Answer:

$w_{Ar}=0.814$

Explanation:

Hello,

In this case, given the temperature, volume and total pressure, we can compute the total moles by using the ideal gas equation:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.040atm*10.0L}{0.082\frac{atm*L}{mol*K}*318K}\\ \\n=0.41mol$

Next, using the molar fraction of argon, we compute the moles of argon:

$n_{Ar}=0.41mol*0.715=0.29mol$

And the moles of methane:

$n_{CH_4}=0.41mol-0.29mol=0.12mol$

Now, by using the molar masses of both argon and methane, we can compute the mass percent of argon:

$w_{Ar}=\frac{m_{Ar}}{m_{Ar}+m_{CH_4}}=\frac{0.29mol*\frac{40g}{1mol} }{0.29mol*\frac{40g}{1mol}+0.12mol*\frac{16g}{1mol}} \\\\w_{Ar}=0.814$

Regards.