The total pressure of a mixture containing equal weights of two non-reacting gases with molecular we — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

Let the weight be $w$. Moles $n_1 = w/4$, $n_2 = w/40$. Total moles = $11w/40$. Mole fraction of lighter gas $X_1 = (w/4) / (11w/40) = 10/11$. Partial pressure $p_1 = X_1 \times P_{total} = (10/11) \times 1.1\text{ atm} = 1.0\text{ atm}$.