Question

At 273 K, the density of a certain gaseous oxide at 2 atmosphere is same as that of dioxygen at 5 atmosphere. The molecular mass of the oxide (in g mol\textsuperscript{--1}) is

• A. 80
• B. 64
• C. 32
• D. 160
• E. 70

Hint:

If density and temperature are constant, the molar mass is inversely proportional to pressure ($M \propto 1/P$). Higher pressure requires a smaller molar mass to maintain the same density!

Answer 1

The Correct Option is A

Concept: The relationship between the density of an ideal gas and its molecular mass is derived from the Ideal Gas Law.
• Density Formula: $d = \frac{PM}{RT}$, where $P$ is pressure, $M$ is molar mass, $R$ is the gas constant, and $T$ is temperature.
• Dioxygen (O\textsubscript{2}): The molar mass of dioxygen is $32 \text{ g mol}^{-1}$.

Step 1: Set up the equality for densities. Given that at constant temperature $T = 273 \text{ K}$, the density of the oxide ($d_1$) equals the density of dioxygen ($d_2$): \[ \frac{P_1 M_1}{RT} = \frac{P_2 M_2}{RT} \] Since $R$ and $T$ are the same for both, they cancel out: \[ P_1 M_1 = P_2 M_2 \]

Step 2: Calculate the molecular mass of the oxide ($M_1$). Substitute the given values: $P_1 = 2 \text{ atm}$, $P_2 = 5 \text{ atm}$, and $M_2 = 32 \text{ g mol}^{-1}$. \[ 2 \times M_1 = 5 \times 32 \] \[ 2M_1 = 160 \] \[ M_1 = \frac{160}{2} = 80 \text{ g mol}^{-1} \]