Question

gs The composition of a gas mixture is 20% O2, 20% N2 and 60% He by mass analysis. What is the molar mass of this mixture

Answer 1

`M_(mix)=3.92(g)/(mol)`

Step-by-step explanation:

Hello!

In this case, since the molar mass of a gas mixture can be computed as a weighted average containing mole fractions:

`M_(mix)=M_(O_2)*x_(O_2)+M_(N_2)*x_(N_2)+M_(He)*x_(He)`

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:

`n_(O_2)=0.20g*(1mol)/(32g)=0.00625mol\\\\ n_(N_2)=0.20g*(1mol)/(28.02)=0.00714mol\\\\n_(He)=0.60g*(1mol)/(4.00)=0.15mol`

Next, we compute the mole fractions:

`x_(O_2)=(0.00625)/(0.00625+0.00714+0.15)=0.038\\\\ x_(O_2)=(0.00714)/(0.00625+0.00714+0.15)=0.044\\\\x_(He)=(0.15)/(0.00625+0.00714+0.15)=0.918`

Now, we compute the molar mass of the mixture as follows:

`M_(mix)=32.0(g)/(mol)* 0.0038+28.02(g)/(mol)*0.0044+4.00(g)/(mol)*0.918\\\\M_(mix)=3.92(g)/(mol)`

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