Question

4- av pusi Identify the unknown diatomic gas that effuses as fast as CO(g) under the same temperature and pressure conditions.

Answer 1

from the Graham's Law of effusion , we know that, Rate of effusion of a gas a . Molecular Mass of that gas ( V MX2 Therefore, if we consider the unknown diatomic gas to be Xz , we can say that Rate of effusion of X2 - Meo Rate of effusion of co Now, as per question, Rate of effusion of X₂ = Rate of effusion of to ii McD = 1 - Мо - Mх, Now, the molecular mass of 20=28.01 Ngay a/mol A known diatomic molecule with same molecular mass in Na gas. Hence, we can say that the unknown diatomic gas is N2 ( Molecular massof) N2 = 28.01 g/ moi ?