To determine the ratio of the densities of the two non-reactive monoatomic ideal gases, we can use the ideal gas law and the given information. The ideal gas law is given by:

$$ PV = nRT $$

Where:

• P is the pressure
• V is the volume
• n is the number of moles
• R is the gas constant
• T is the temperature

Given that the gases are monoatomic and non-reactive, their molecular masses are directly proportional to their atomic masses. Let the atomic masses of the two gases be $M_1$ and $M_2$, with $M_1 : M_2 = 2 : 3$.

The density ($\rho$) of a gas is given by the formula:

$$ \rho = \frac{m}{V} $$

Where $m$ is the mass and $V$ is the volume. Using the ideal gas law, we can relate density to pressure and temperature.

From the ideal gas law:

$$ PV = nRT $$

Since $ n = \frac{m}{M} $, where $M$ is the molar mass, we have:

$$ P = \frac{mRT}{MV} $$

Rearranging to solve for density ($\rho\ = \frac{m}{V}$):

$$ \rho = \frac{PM}{RT} $$

Thus, the density of a gas is directly proportional to its pressure and molar mass, and inversely proportional to the temperature. Given that the ratio of their partial pressures $P_1 : P_2 = 4 : 3$, we can write:

$$ \rho_1 = \frac{P_1 M_1}{RT} $$

$$ \rho_2 = \frac{P_2 M_2}{RT} $$

Taking the ratio of the densities:

$$ \frac{\rho_1}{\rho_2} = \frac{\left(\frac{P_1 M_1}{RT}\right)}{\left(\frac{P_2 M_2}{RT}\right)} = \frac{P_1 M_1}{P_2 M_2} $$

Substituting the given ratios, $ P_1 : P_2 = 4 : 3 $ and $ M_1 : M_2 = 2 : 3 $, into the equation, we get:

$$ \frac{\rho_1}{\rho_2} = \frac{4 \times 2}{3 \times 3} = \frac{8}{9} $$

Therefore, the ratio of their densities is $ \frac{8}{9} $, which corresponds to option D. Consequently, the answer is:

Option D: 8 : 9