Question

1. a gas mixture contains 1.25 g n₂ and 0.85 g o₂ in a 1.55 l container at 18°c. calculate the mole fraction and partial pressure of each component in the gas mixture.

Explanation:

Step1: Calculate the number of moles of \(N_2\) and \(O_2\)

The molar mass of \(N_2\) is \(M_{N_2}=28\space g/mol\). Using the formula \(n=\frac{m}{M}\), for \(N_2\), \(n_{N_2}=\frac{1.25\space g}{28\space g/mol}\approx0.0446\space mol\).
The molar mass of \(O_2\) is \(M_{O_2} = 32\space g/mol\). For \(O_2\), \(n_{O_2}=\frac{0.85\space g}{32\space g/mol}\approx0.0266\space mol\).

Step2: Calculate the total number of moles

Using the formula \(n_{total}=n_{N_2}+n_{O_2}\), \(n_{total}=0.0446 + 0.0266=0.0712\space mol\)

Step3: Calculate the mole fraction

The mole - fraction formula is \(x_i=\frac{n_i}{n_{total}}\).
For \(N_2\), \(x_{N_2}=\frac{0.0446}{0.0712}\approx0.626\)
For \(O_2\), \(x_{O_2}=\frac{0.0266}{0.0712}\approx0.374\)

Step4: Convert the temperature to Kelvin

Using the formula \(T(K)=T(^{\circ}C)+ 273.15\), \(T = 18 + 273.15=291.15\space K\)

Step5: Calculate the total pressure using the ideal gas law \(PV=nRT\)

\(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(V = 1.55\space L\), \(n = 0.0712\space mol\), \(T=291.15\space K\)
\(P=\frac{nRT}{V}=\frac{0.0712\times0.0821\times291.15}{1.55}\)
\(P=\frac{0.0712\times0.0821\times291.15}{1.55}\approx1.09\space atm\)

Step6: Calculate the partial pressure

Using the formula \(P_i=x_iP\)
For \(N_2\), \(P_{N_2}=x_{N_2}P=0.626\times1.09\approx0.682\space atm\)
For \(O_2\), \(P_{O_2}=x_{O_2}P=0.374\times1.09\approx0.408\space atm\)

Answer:

• Mole fraction of \(N_2\): \(x_{N_2}\approx0.626\), mole fraction of \(O_2\): \(x_{O_2}\approx0.374\)
• Partial pressure of \(N_2\): \(P_{N_2}\approx0.682\space atm\), partial pressure of \(O_2\): \(P_{O_2}\approx0.408\space atm\)