Question

Match List-I (Atomic masses M) with List-II (% composition of heavier isotope).

Answer 1

As we know,
$Avg.mass=\frac{x}{100}$
Also,
$\%$ by moles $=\frac{M_{avg}-M_1}{M_2-M_1}\times 100$
$\%$ by mass$=\%$ by moles $\times \frac{M_2}{M_{avg}}$
(A) $\%$ by moles $=1\times 100/4=25\%$.
(B)$\%$ by moles $=1\times 100/2=50\%$.
(C) $\%$ by mass $=50\times 3/2=75\%$.