Question

a mixture of three gases has a total pressure of 1380 mm hg at 298 k. the mixture is analysed and is found to contain 1.27 mol carbon dioxide, 3.04 mol carbon monoxide, and 1.50 mol argon. what is the partial pressure of argon? 301 mm hg 5345 mm hg 8020 mm hg 356 mm hg 0.258 mm hg

Explanation:

Step1: Calculate the total number of moles

The total number of moles \(n_{total}=n_{CO_2}+n_{CO}+n_{Ar}\).
Substitute \(n_{CO_2} = 1.27\space mol\), \(n_{CO}=3.04\space mol\), \(n_{Ar}=1.50\space mol\) into the formula:
\(n_{total}=1.27 + 3.04+1.50=5.81\space mol\)

Step2: Calculate the mole fraction of argon

The mole fraction of argon \(X_{Ar}=\frac{n_{Ar}}{n_{total}}\)
Substitute \(n_{Ar} = 1.50\space mol\) and \(n_{total}=5.81\space mol\) into the formula:
\(X_{Ar}=\frac{1.50}{5.81}\approx0.258\)

Step3: Calculate the partial pressure of argon

According to Dalton's law of partial pressures \(P_{Ar}=X_{Ar}\times P_{total}\)
Substitute \(X_{Ar}\approx0.258\) and \(P_{total} = 1380\space mm\space Hg\) into the formula:
\(P_{Ar}=0.258\times1380 = 356\space mm\space Hg\)

Answer:

356 mm Hg