Question

when u answer can u please write clear and explain in detail. thank you

2. Natural rubidium has the average mass of 85.4678 g/mol and is composed of isotopes Rb (mass = 84.9117 g/mol) and 7 Rb. The ratio of atoms *S Rb/87 Rb in natural rubidium is 2.591. Calculate the mass of Rb. Numerical solution: (1) C1-35 75.76%; (2) 86.9085 g/mol

Answer 1

Let atomic mass of ⁸⁷Rb = x

ratio of  ⁸⁵Rb and ⁸⁷Rb is 2.591

hence fraction of mass contributed by ⁸⁵Rb = 2.591

and fraction of mass contributed by ⁸⁷Rb = 1

now, atomic mass of Rb =

fraction of mass contributed by S7 Rb x2 + fraction of mass contributed by Rb x 84.9117 total fraction contributed

or, 85.4678 =    

1x2 +2.591 X 84.9117 1+2.591

or , 85.4678 * (1+2.591) = x + 222.0062

or, 306.9147 = x + 220.0062

or, x = 306.9147  - 220.0062

or, x = 86.9085 g/mol.

so atomic mass of ⁸⁷Rbis 86.9085 g/mol