Boron exists as two stable isotopes; ¹⁰B(19%) and ¹¹B(81%). Find out the average atomic weight of boron in the periodic table

The correct option is 10.8.

To find the average atomic weight of boron, we calculate the weighted average of the atomic masses of its stable isotopes based on their relative natural abundance. The two isotopes and their abundances are given as:
- ¹⁰B with an abundance of 19% (or 0.19 as a fraction)
- ¹¹B with an abundance of 81% (or 0.81 as a fraction)

The formula for the average atomic weight is:
$$\text{Average Atomic Weight}=({\text{Mass}}_1\times {\text{Abundance}}_1)+({\text{Mass}}_2\times {\text{Abundance}}_2)$$

Substituting the values of the isotopes into the formula:
$$\text{Average Atomic Weight}=(10\times \frac{19}{100})+(11\times \frac{81}{100})$$

Now, we perform the individual multiplications:
$$\text{Average Atomic Weight}=(10\times 0.19)+(11\times 0.81)$$

Calculating the terms gives:
$$\text{Average Atomic Weight}=1.9+8.91$$

Adding the two values together:
$$\text{Average Atomic Weight}=10.81$$

Rounding to one decimal place as presented in the options, we get 10.8.