Chemistry - Mole Concept Question with Solution | TestHub

The correct option is (D).

(A) $\to$ (Q): An equimolar mixture of O₃ ($M=48$) and O₂ ($M=32$) gives $M_{avg}=\frac{48+32}{2}=40$.

(B) $\to$ (R): For a $3:1$ molecular ratio of CH₄ (16) and SO₃ (80), $M_{avg}=\frac{3(16)+1(80)}{3+1}=32$.

(C) $\to$ (P): S:O = 1:5 implies SO₃ and O₂ are equimolar ($x=y$), so $M_{avg}=\frac{80+32}{2}=56$.

(C) $\to$ (S): For a $1:27$ mass ratio of He and N₂O₅, $M_{avg}=\frac{TotalMass}{TotalMoles}=\frac{1+27}{\frac{1}{4}+\frac{27}{108}}=\frac{28}{0.5}=56$.