At the sea level, the dry air mass percentage composition is given as nitrogen gas : 70.0, oxygen gas : 27.0 and argon gas : 3.0. If total pressure is 1.15 atm, then calculate the ratio of followings respectively :
(i) partial pressure of nitrogen gas to partial pressure of oxygen gas
(ii) partial pressure of oxygen gas to partial pressure of argon gas
(Given : Molar mass of $$N, O$$ and $$Ar$$ are 14, 16, and 40 g $$mol^{-1}$$ respectively)

For an ideal gas mixture, the partial pressure $$P_i$$ of any component is given by
$$P_i = x_i \, P_{\text{total}}$$
where $$x_i$$ is the mole fraction of that component. Hence, the ratio of two partial pressures equals the ratio of the corresponding mole fractions.

The composition is provided as mass % at sea level:
nitrogen (N2) : 70.0 %, oxygen (O2) : 27.0 %, argon (Ar) : 3.0 %.

Assume 100 g of this dry air (any convenient mass may be chosen).
  Mass of N2 = 70 g  Mass of O2 = 27 g  Mass of Ar = 3 g

Convert each mass to moles using molar masses $$M$$:
$$n_{\text{N}_2} = \frac{70}{28} = 2.50 \text{ mol}$$
$$n_{\text{O}_2} = \frac{27}{32} = 0.84375 \text{ mol}$$
$$n_{\text{Ar}} = \frac{3}{40} = 0.075 \text{ mol}$$

Total moles
$$n_{\text{total}} = 2.50 + 0.84375 + 0.075 = 3.41875 \text{ mol}$$

Mole fractions
$$x_{\text{N}_2} = \frac{2.50}{3.41875} = 0.7313$$
$$x_{\text{O}_2} = \frac{0.84375}{3.41875} = 0.2467$$
$$x_{\text{Ar}} = \frac{0.075}{3.41875} = 0.02193$$

Because $$\dfrac{P_{\text{A}}}{P_{\text{B}}} = \dfrac{x_{\text{A}}}{x_{\text{B}}}$$ :

Case (i):
$$\frac{P_{\text{N}_2}}{P_{\text{O}_2}} = \frac{x_{\text{N}_2}}{x_{\text{O}_2}} = \frac{0.7313}{0.2467} \approx 2.96$$

Case (ii):
$$\frac{P_{\text{O}_2}}{P_{\text{Ar}}} = \frac{x_{\text{O}_2}}{x_{\text{Ar}}} = \frac{0.2467}{0.02193} \approx 11.2$$

Thus the required ratios are
(i) $$2.96$$ and (ii) $$11.2$$.

Comparing with the options, these correspond to Option D.