Question

# Air contains $23%$ of oxygen and $77%$of nitrogen by weight. What is the percentage of oxygen by volume is?

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Solution

## Step 1: Given dataPercentage of oxygen$=23%$Percentage of nitrogen$=77%$Step 2: Calculating the percentage of oxygenLet the mass of air$=100\mathrm{g}$Therefore, the mass of oxygen in $100\mathrm{g}$ of air$=23\mathrm{g}$and mass of nitrogen in $100\mathrm{g}$ of air$=77\mathrm{g}$Also, the molecular mass of oxygen$=32\mathrm{g}$and the molecular mass of nitrogen$=28\mathrm{g}$Now, the number of moles of oxygen$=\frac{\mathrm{given}\mathrm{mass}}{\mathrm{molecular}\mathrm{mass}}$$⇒\frac{22\mathrm{g}}{32\mathrm{g}}$and number of moles of nitrogen$=\frac{\mathrm{given}\mathrm{mass}}{\mathrm{molecular}\mathrm{mass}}$$⇒\frac{77\mathrm{g}}{28\mathrm{g}}$As one mole of gas requires $22.4\mathrm{L}$ of volume at Standard Temperature and Pressure (STP) conditions, SoThe volume occupied by moles of oxygen$=\frac{23\mathrm{g}}{32\mathrm{g}}×22.4\mathrm{L}⇒16.1\mathrm{L}$The volume occupied moles of nitrogen$=\frac{77\mathrm{g}}{28\mathrm{g}}×22.4\mathrm{L}⇒61.6\mathrm{L}$Therefore, the total volume of air$=$volume occupied by moles of oxygen$+$volume occupied by moles of nitrogen$=16.6\mathrm{L}+61.6\mathrm{L}⇒77.7\mathrm{L}$So percentage by volume of oxygen$=\frac{\mathrm{volume}\mathrm{occupied}\mathrm{by}\mathrm{moles}\mathrm{of}\mathrm{oxygen}}{\mathrm{total}\mathrm{volume}\mathrm{of}\mathrm{air}}×100$$=\frac{16.1}{77.7}×100⇒20.7%$Therefore, the percentage of oxygen by volume is $20.7%$

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