Let x be the natual abundance of lower mass Since B has only two isotopes so the abundance of heavear mass is 1 - x The average atomic mass = [(mass of lower isotope * its abundance) + ( mass of heavier mass* itys abundance)] / no.of isotopes 10.845 = [(10.013* x ) + ( 11.009*(1-x) ] / 2 ---> x = 0.508 % abundance = 0.508*100 = 50.8 % Solution Let x be the natual abundance of lower mass Since B has only two isotopes so the abundance of heavear mass is 1 - x The average atomic mass = [(mass of lower isotope * its abundance) + ( mass of heavier mass* itys abundance)] / no.of isotopes 10.845 = [(10.013* x ) + ( 11.009*(1-x) ] / 2 ---> x = 0.508 % abundance = 0.508*100 = 50.8 %.