An unknown gas 'X' has rate of diffusion measured to be 0.88 times that of PH3 at the same condition — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

r_X / r_PH3 = sqrt[M(PH3)/M_X] => 0.88 = √(34/M_X) => M_X = 34 / 0.774 = 43.9 g/mol ≈ 44 g/mol . Since molecular weight of $N_2O$ is 44, the gas is $N_2O$ .