QJEEAdv78Ch4424: JEE Advanced 1978 Chemistry - Class 11 JEEAdv78Ch4424 Solution: Naturally occurring boron consists of two isotopes whose atomic …

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%