A) 19 : 81
B) 10 : 11
C) 15 : 61
D) 81 : 19
Correct Answer: A
Solution :
Let percentage of\[{{B}^{10}}\]atom is\[x,\]then average atomic weight \[=\frac{10x+11(100-x)}{100}\] \[=10.81\] \[x=19\] \[\therefore \] \[\frac{_{5}{{B}^{10}}}{_{5}{{B}^{11}}}=\frac{19}{81}\]You need to login to perform this action.
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