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