Question

a mixture with h₂ and he exerts a total pressure of 0.48 atm. if there is 1.0 g of h₂ and 1.0 g of he in the mixture, what is the partial pressure of the helium gas?
select the correct answer below:
0.48 atm
0.32 atm
0.16 atm
0.08 atm

Explanation:

Step1: Calculate moles of each gas

Molar mass of \(H_2\) is \(M_{H_2}=2\space g/mol\), moles of \(H_2\), \(n_{H_2}=\frac{m}{M}=\frac{1.0\space g}{2\space g/mol} = 0.5\space mol\)
Molar mass of \(He\) is \(M_{He}=4\space g/mol\), moles of \(He\), \(n_{He}=\frac{m}{M}=\frac{1.0\space g}{4\space g/mol}=0.25\space mol\)

Step2: Calculate mole fraction of \(He\)

Mole fraction of \(He\), \(x_{He}=\frac{n_{He}}{n_{H_2}+n_{He}}=\frac{0.25}{0.5 + 0.25}=\frac{0.25}{0.75}=\frac{1}{3}\)

Step3: Calculate partial pressure of \(He\)

According to Dalton's law of partial pressures \(P_{He}=x_{He}\times P_{total}\)
Given \(P_{total} = 0.48\space atm\)
\(P_{He}=\frac{1}{3}\times0.48\space atm = 0.16\space atm\)

Answer:

0.16 atm