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timofeeve [1]
3 years ago
7

I'll be giving out brainliest to whoever chooses more than one!!!

Chemistry
1 answer:
Kruka [31]3 years ago
8 0

I think it might be B, and A

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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
andrey2020 [161]

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.

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