A) \[19:81\]
B) \[10: 11\]
C) \[15 : 16\]
D) \[81 : 19\]
Correct Answer: A
Solution :
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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