Question

A 13.0-L scuba diving tank contains a helium-oxygen (heliox) mixture made up of 23.6 g of He and 4.85 g of O2 at 298 K. Calculate the mole fraction of each component in the mixture.

Answer 1

Answer:

$x_{He}=0.975\\x_{O_2}=0.025$

Explanation:

Hello.

In this case, since we know the mass of both helium and oxygen, in order to obtain the mole fractions we first need the compute the moles by using their atomic masses, 4.00 g/mol and 32.00 g/mol respectively as shown below:

$n_{He}=23.6gHe*\frac{1molHe}{4.00gHe}=5.90molHe\\ \\n_{O_2}=4.85gO_2*\frac{1molO_2}{32.00gO_2}=0.152molO_2$

Therefore, the mole fractions are:

$x_{He}=\frac{n_{He}}{n_{He}+n_{O_2}}=\frac{5.90}{5.90+0.152} \\\\x_{He}=0.975\\\\x_{O_2}=\frac{n_{O_2}}{n_{He}+n_{O_2}} =\frac{0.152}{5.90+0.152} \\\\x_{O_2}=0.025$

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