1.

Calculatethe %relativeabundanceofB-10and B-11, ifitsaverage atomicmassis 10 .804amu .

Answer»

SOLUTION :Let%Abundanceof `B-10 = X `
%of relativeabundanceof `B - 11 =(100 - x)`
Averageatomicmassof B `=(10xx x+11 (100-x))/(100)`
`10.804 =(10x +1100- 11 x )/( 100 )`
`10.804 xx 100 = 1100 - x `
`1080.4 = 1100-x `
` x=1100 - 1080.4 `
` thereforex= 19.6 % `
` THEREFORE ` %abundanceof b- 10is 19.6 %
` therefore ` % Abundanceof B-11= (100 -x)
`=100-19.6 `
`=80.4`


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