The atomic weight of boron is 10.81 and it ha | vedclass

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(A) Let the percentage of $_5B^{10}$ atoms be $x$. Then the percentage of $_5B^{11}$ atoms is $(100 - x)$.The average atomic weight is given by the fo

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$19 : 81$

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(A) Let the percentage of $_5B^{10}$ atoms be $x$. Then the percentage of $_5B^{11}$ atoms is $(100 - x)$.The average atomic weight is given by the formula:$	ext{Average Atomic Weight} = \frac{(10 	imes x) + (11 	imes (100 - x))}{100} = 10.81$Multiplying by $100$:$10x + 1100 - 11x = 1081$$-x = 1081 - 1100$$-x = -19$$x = 19$So,the percentage of $_5B^{10}$ is $19\%$ and the percentage of $_5B^{11}$ is $81\%$.Therefore,the ratio of $_5B^{10} : _5B^{11}$ is $19 : 81$.

The atomic weight of boron is $10.81$ and it has two isotopes $_5B^{10}$ and $_5B^{11}$. The ratio of $_5B^{10} : _5B^{11}$ in nature is:

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