1. Given mass percentage composition of dry air

Assume we have $100\,\text{g}$ of dry air.

• Nitrogen gas: $70\,\text{g}$
• Oxygen gas: $27\,\text{g}$
• Argon gas: $3\,\text{g}$

2. Convert masses to moles

For nitrogen gas, $\mathrm{N_2}$: $M(\mathrm{N_2}) = 2\times 14 = 28\,\text{g mol}^{-1}$ $n_{\mathrm{N_2}} = \frac{70}{28} = 2.5$

For oxygen gas, $\mathrm{O_2}$: $M(\mathrm{O_2}) = 2\times 16 = 32\,\text{g mol}^{-1}$ $n_{\mathrm{O_2}} = \frac{27}{32} = 0.84375$

For argon gas, $\mathrm{Ar}$: $M(\mathrm{Ar}) = 40\,\text{g mol}^{-1}$ $n_{\mathrm{Ar}} = \frac{3}{40} = 0.075$

3. Use Dalton's law of partial pressure

For an ideal gas mixture, $p_i = x_i P$ where $x_i$ is mole fraction.

Hence, ratios of partial pressures are same as ratios of mole numbers: $\frac{p_{\mathrm{N_2}}}{p_{\mathrm{O_2}}} = \frac{n_{\mathrm{N_2}}}{n_{\mathrm{O_2}}}$ $\frac{p_{\mathrm{O_2}}}{p_{\mathrm{Ar}}} = \frac{n_{\mathrm{O_2}}}{n_{\mathrm{Ar}}}$

4. Calculate the required ratios

(i) Ratio of partial pressure of nitrogen to oxygen: $\frac{p_{\mathrm{N_2}}}{p_{\mathrm{O_2}}} = \frac{2.5}{0.84375} = 2.96296 \approx 2.96$

(ii) Ratio of partial pressure of oxygen to argon: $\frac{p_{\mathrm{O_2}}}{p_{\mathrm{Ar}}} = \frac{0.84375}{0.075} = 11.25 \approx 11.2$

5. Match with options

The pair is: $2.96,\ 11.2$

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They match.