The vapour density of an equilibrium mixture of and gases is experimentally found to be exactly at . — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

Step 1: Calculate average molar mass of the mixture. $M_{\text{avg}} = 2 \times V.D. = 2 \times 39 = 78\text{ g/mol}$. Step 2: Let the mass of $NO_2$ be $x$ g, so $N_2O_4$ is $(100-x)$ g. Step 3: Apply the average molar mass formula: $M_{\text{avg}} = \frac{\text{Total Mass}}{\text{Total Moles}}$. $78 = \frac{100}{\frac{x}{46} + \frac{100-x}{92}}$. Solving: $\frac{100}{78} = \frac{2x + 100 - x}{92} \implies 9200 = 78(100 + x) \implies 9200 = 7800 + 78x \implies 1400 = 78x \implies x = 17.9\text{ g}$.