**SOLUTION : **Suppose the percentage of ${ }^{12} C$ isotope of mass $12 \cdot 0000=x$ and

hence that of ${ }^{13} C$ isotope of mass $13.0034=(100-x)$

$ \text { Atomic mass of carbon }=\frac{x \times 12 \cdot 0+(100-x) \times 13 \cdot 0034}{100}$

$\therefore 12 \cdot 0111=\frac{12 x+1300 \cdot 34-13 \cdot 0034 x}{100}$

$\text { or, } \quad 1201 \cdot 11=-1 \cdot 0034 x+1300 \cdot 34,1 \cdot 0034 x=1300 \cdot 34-1201 \cdot 11 =99 \cdot 23$

$\therefore \therefore \text { and } \% \text { of isotope of mass } 13 \cdot 0034= 1 \cdot 1 0 6 \% .$