The ratio of $\mathrm{N}_2$ and $\mathrm{O}_2$ gases in the atmosphere is $4: 1$. The ratio of the mole fractions of the dissolved gases $\mathrm{N}_2: \mathrm{O}_2$ in rain water will be approximately ------- (At $293 \mathrm{~K}, \mathrm{~K}_{\mathrm{H}}$ for Nitrogen and Oxygen in kbar units are 76.48 and 34.86 respectively.)

According to Henry's Law, the solubility of a gas in a liquid is given by $P = K_{H} \times x$, where $P$ is the partial pressure of the gas, $K_{H}$ is the Henry's law constant, and $x$ is the mole fraction of the gas in the liquid.

The partial pressure of a gas is proportional to its mole fraction in the gas phase. Given the ratio of $\mathrm{N}_{2}$ and $\mathrm{O}_{2}$ in the atmosphere is $4:1$, we have $P_{\mathrm{N}_{2}} = 4P$ and $P_{\mathrm{O}_{2}} = 1P$ for some constant $P$.

The mole fractions of the dissolved gases are $x_{\mathrm{N}_{2}} = \dfrac{P_{\mathrm{N}_{2}}}{K_{H, \mathrm{N}_{2}}}$ and $x_{\mathrm{O}_{2}} = \dfrac{P_{\mathrm{O}_{2}}}{K_{H, \mathrm{O}_{2}}}$.

The ratio of the mole fractions in water is $\dfrac{x_{\mathrm{N}_{2}}}{x_{\mathrm{O}_{2}}} = \dfrac{P_{\mathrm{N}_{2}}}{K_{H, \mathrm{N}_{2}}} \times \dfrac{K_{H, \mathrm{O}_{2}}}{P_{\mathrm{O}_{2}}}$.

Substituting the given values: $\dfrac{x_{\mathrm{N}_{2}}}{x_{\mathrm{O}_{2}}} = \dfrac{4}{1} \times \dfrac{34.86}{76.48}$.

Calculating the value: $\dfrac{x_{\mathrm{N}_{2}}}{x_{\mathrm{O}_{2}}} = 4 \times 0.4558 \approx 1.823$.

This value is approximately $2:1$.

Answer: $2: 1$