JEE Advance - Chemistry (1978 - No. 3)
Explanation
Let, percentage of isotope with atomic weight $$10.01 = x$$
and percentage of isotope with atomic weight $$11.01 = 100 - x$$
We know, average atomic weight formula,
$${A_{avg}} = {{\sum {\left( {\% \,\mathrm{of}\,{I_i} \times \mathrm{Atomic\,Weight}} \right)} } \over {\sum {\left( {\% \,\mathrm{of}\,{I_i}} \right)} }}$$
Here $${I_i}$$ = i th isotope of an element.
Here given,
$${A_{avg}} = 10.81$$
$$\therefore$$ $$10.81 = {{10.01 \times x + 11.01 \times (100 - x)} \over {x + 100 - x}}$$
$$ \Rightarrow 10.81 = {{10.01x + 11.01(100 - x)} \over {100}}$$
$$ \Rightarrow 1081 = 10.01x + 1101 - 11.01x$$
$$ \Rightarrow x = 1101 - 1081$$
$$ \Rightarrow x = 20$$
$$\therefore$$ Isotope with weight 10.01 present = 20% and isotope with weight 11.01 present = 80%
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