A) 19 : 81
B) 10 : 11
C) 15 : 16
D) 81 : 19
Correct Answer: A
Solution :
[a] Let the percentage of \[{{B}^{10}}\] atoms be x, then average atomic weight \[=\frac{10x+11(100-x)}{100}=10.81\Rightarrow x=19\] \[\therefore \,\,\,\,\,\,\,\,\,\frac{{{N}_{{{B}^{10}}}}}{{{N}_{{{B}^{11}}}}}=\frac{19}{81}\]You need to login to perform this action.
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