Question

Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.538 g/L at 25 ∘C and 721 mmHg?

Answer 1

Answer: The mole percentage of helium in the mixture is 64.75 %

Explanation:

To calculate the molar mass of mixture, we use ideal gas equation, which is:

$PV=nRT$

Or,

$P=\frac{m}{M}\frac{RT}{V}$

We know that:

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

Rearranging the above equation:

$M=\frac{dRT}{P}$

where,

M = molar mass of mixture = ?

d = density of mixture = 0.538 g/L

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

P = pressure of the mixture = 721 mmHg

Putting values in above equation, we get:

$M=\frac{0.538g/L\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K}{721mmHg}\\\\M=13.87g/mol$

Molar mass of the mixture will be the sum of molar mass of each substance each multiplied by its mole fraction.

Let the mole fraction of Helium be 'x' and that of oxygen be '1-x'

$M=(x\times M_{He})+((1-x)\times M_{O_2})$

We know that:

Molar mass of helium = 4.00 g/mol

Molar mass of oxygen gas = 32g/mol

Putting values in above equation, we get:

$13.87=(x\times 4)+((1-x)\times 32)\\\\x=0.6475$

Mole fraction of helium in the mixture = 0.6475

Calculating the mole percentage of helium in the mixture:

$\text{Mole percentage of helium in the mixture}=x\times 100\\\\\text{Mole percentage of helium in the mixture}=0.6475\times 100=64.75\%$

Hence, the mole percentage of helium in the mixture is 64.75 %