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JEE Advance - Chemistry (1978 - No. 3)

Naturally occurring boron consists of two isotopes whose atomic weighs are 10.01 and 11.01. The atomic weight of natural oron is 10.81. Calculate the percentage of each isotope in natural boron.
10%, 90%
20%, 80%
30%, 70%
40%, 60%
50%, 50%

Selitys

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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