The relative atomic mass is 10.8amu. What is the % of "1 of boron which consists of the isotopes "5B, $B of atoms in the isotopic mixture?

Explanation

Boron has two main isotopes: Boron-10 (⁵B¹⁰) and Boron-11 (⁵B¹¹). The relative atomic mass of boron, which is the weighted average of the masses of its isotopes, is given as 10.8 amu. We need to find the percentage of Boron-10 (⁵B¹⁰) in the isotopic mixture.Here's how to solve this problem:Let x represent the abundance of Boron-10 (⁵B¹⁰) as a decimal.Set up an equation using the weighted average mass formula:10.8 amu = (x * 10.013 amu) + ((1 - x) * 11.009 amu)10.013x + 11.009 - 11.009x = 10.8-0.996x = -0.2x = 0.2008 (round to four decimal places)Convert the decimal abundance to a percentage:Percentage of ⁵B¹⁰ = x * 100%Percentage of ⁵B¹⁰ = 0.2008 * 100% = 20.08% (round to one decimal place)Therefore, approximately 20.1% of the boron atoms in the mixture are Boron-10 (⁵B¹⁰).