Assuming ideal gas behavior, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is: (Atomic weight of Cl is 35.5 u)

For any ideal gas, the density $$\rho$$ at a given temperature $$T$$ and pressure $$P$$ is related to its molar mass $$M$$ by the ideal-gas formula rearranged for density

$$PV = nRT \;\;\Longrightarrow\;\; P = \dfrac{nRT}{V} = \dfrac{\left(\dfrac{m}{M}\right)RT}{V} = \dfrac{mRT}{MV}$$

Dividing both sides by $$RT$$, we get

$$\dfrac{P}{RT} = \dfrac{m}{MV} = \dfrac{\rho}{M}$$  because $$\rho = \dfrac{m}{V}$$.

So, explicitly,

$$\rho = \dfrac{P}{RT}\,M.$$

We have the same $$P$$ and the same $$T$$ for both gases, therefore the factor $$\dfrac{P}{RT}$$ is common and cancels out when we take a ratio. Hence, the ratio of densities equals the ratio of molar masses:

$$\dfrac{\rho_{\text{NH}_3}}{\rho_{\text{HCl}}} = \dfrac{M_{\text{NH}_3}}{M_{\text{HCl}}}.$$

Now we calculate each molar mass step by step.

For ammonia, NH3:

$$M_{\text{NH}_3} = M_{\text{N}} + 3M_{\text{H}} = 14.0 \,\text{u} + 3(1.0 \,\text{u}) = 14.0 \,\text{u} + 3.0 \,\text{u} = 17.0 \,\text{u}.$$

For hydrogen chloride, HCl:

Atomic mass of H = 1.0 u (given implicitly), atomic mass of Cl = 35.5 u (given).

$$M_{\text{HCl}} = M_{\text{H}} + M_{\text{Cl}} = 1.0 \,\text{u} + 35.5 \,\text{u} = 36.5 \,\text{u}.$$

Substituting these two molar masses into the ratio expression, we obtain

$$\dfrac{\rho_{\text{NH}_3}}{\rho_{\text{HCl}}} = \dfrac{17.0}{36.5}.$$

Now we carry out the division numerically:

$$\dfrac{17.0}{36.5} = 0.46575\ldots \approx 0.46.$$

So, the density of ammonia is about 0.46 times that of hydrogen chloride at the same temperature and pressure.

Hence, the correct answer is Option C.