Question
10.A sample of boron is made up of 18.9% of boron -10 atoms and 81.1% of boron -11 atoms. What is the relative atomic mass of boron? Calculate to 2 decimal places
Answer
3.6
(161 Votes)
Georgia
Master · Tutor for 5 years
Answer
10.81
Explanation
The question asks us to calculate the relative atomic mass of Boron given the percentage composition of its isotopes. Relative atomic mass is the weighted average of the atomic masses of the naturally-occurring isotopes of an element. It is calculated using the formula:Relative atomic mass (´㏒/mol)= Σ(isotope% x isotope mass)/100The two isotopes of boron are Boron-10 and Boron-11, which have respective masses of approximately 10 amu and 11 amu. The question gives us the isotopic distribution as 18.9% of Boron-10 and 81.1% of Boron-11.Using the provided formula, we get:Weighted atomic mass (% times amu of each isotope summed, then divided by 100)= [(18.9% x 10amu) + (81.1% x 11amu)] / 100= [189 + 892.1] / 100The calculation of which gives us the relative atomic mass to two decimal places