Question

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following.The mole fraction of gas B?

Answer 1

The question is incomplete, here is the complete question:

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?

A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.

Answer: The mole fraction of gas B (hydrogen gas) is 0.557

Explanation:

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

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

• For nitrogen gas:

Given mass of nitrogen gas = 10.25 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen gas}=\frac{10.25g}{28g/mol}=0.366mol$

• For hydrogen gas:

Given mass of hydrogen gas = 2.05 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{2.05g}{2g/mol}=1.025mol$

• For ammonia gas:

Given mass of ammonia gas = 7.63 g

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia gas}=\frac{7.63g}{17g/mol}=0.449mol$

Mole fraction of a substance is given by:

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

Moles of gas B (hydrogen gas) = 1.025 moles

Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles

Putting values in above equation, we get:

$\chi_{(H_2)}=\frac{1.025}{1.84}=0.557$

Hence, the mole fraction of gas B (hydrogen gas) is 0.557