Question

A gas mixture consists of equal masses of methane (molecular weight 16.0) and argon (atomic weight 40.0). If the partial pressure of argon is 200. torr, what is the pressure of methane, in torr

Answer 1

Answer:

80.11 torr

Explanation:

Hello,

In this case, since the masses are equal, we could assume 1 gram for each gas, which result in the following moles:

$n_{CH_4}=\frac{1g}{16.0g/mol}=0.0625molCH_4 \\n_{Ar}=\frac{1g}{40.0g/mol}=0.025molAr$

Thus, the molar fractions:

$x_{CH_4}=\frac{0.0625}{0.0625+0.025}=0.714\\x_{Ar}=1- x_{CH_4}=1-0.714=0.286$

Thus, the total pressure is:

$p=p_{CH_4}+p_{Ar}\\p=x_{CH_4}p+x_{Ar}p\\p=200torr+0.286p\\p=\frac{200torr}{1-0.286}\\ p=280.11torr$

Hence the pressure of methane:

$p_{CH_4}=280.11torr-200torr=80.11torr$

Best regards.