Natural rubidium has the average mass of 85.4678 and is composed of isotopes 85 rb(mass = 84.9117) and 87 rb. the ratio of atoms 85rb/87rb in natural rubidium is 2.591. calculate the mass of 87 rb.

Mass of Rb-87 is 86.913 amu.

Explanation:

Given data:

Average mass of rubidium = 85.4678 amu

Mass of Rb-85 = 84.9117

Ratio of 85Rb/87Rb in natural rubidium = 2.591

Mass of Rb = ?

Solution:

The ration of both isotope is 2.591 to 1. Which means that for 2.591 atoms of Rb-85 there is one Rb-87.

For 100% naturally occurring Rb = 2.591 + 1 = 3.591

% abundance of Rb-85 = 2.591/ 3.591 = 0.722

% abundance of Rb-87 = 1 - 0.722= 0.278

84.9117 × 0.722 + X × 0.278 = 85.4678

61.306 + X × 0.278 = 85.4678

X × 0.278 = 85.4678 - 61.306

X × 0.278 = 24.1618

X =  24.1618 / 0.278

X = 86.913 amu

the child will have heterozygous dominant chromosomes as characters of his father are dominant over the characters of his mother.

the child is heterozygous