Question

A 15.0-L scuba diving tank contains a helium-oxygen (heliox) mixture made up of 23.0 g of He and 4.14 g of O2 at 298 K. Calculate the mole fraction of each component in the mixture.

Answer 1

Answer: The mole fraction of Helium is 0.978 and that of oxygen gas is 0.022

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For helium:

Given mass of helium = 23 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$\text{Moles of helium}=\frac{23g}{4g/mol}=5.75mol$

• For oxygen gas:

Given mass of oxygen gas = 4.14 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{4.14g}{32g/mol}=0.129mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

• For helium:

Moles of helium = 5.75 moles

Total moles = [5.75 + 0.129] = 5.879 moles

Putting values in above equation, we get:

$\chi_{(He)}=\frac{5.75}{5.879}=0.978$

• For oxygen gas:

Moles of oxygen gas = 0.129 moles

Total moles = [5.75 + 0.129] = 5.879 moles

Putting values in above equation, we get:

$\chi_{(O_2)}=\frac{0.129}{5.879}=0.022$

Hence, the mole fraction of Helium is 0.978 and that of oxygen gas is 0.022