A) 19 : 81
B) 10 : 11
C) 15 : 16
D) 81 : 19
Correct Answer: A
Solution :
| Let \[{{n}_{1}}\] and \[{{n}_{2}}\] be the number of atoms in \[_{5}^{10}B\] and \[_{5}^{11}B\] isotopes. |
| Atomic weight |
| \[=\frac{{{n}_{1}}\times (At.\,wt.\,of\,_{5}^{10}B)+{{n}_{2}}\times (At.wt.\,of\,_{5}^{11}B)}{{{n}_{1}}+{{n}_{2}}}\] |
| or \[10.81=\frac{{{n}_{1}}\times 10+{{n}_{2}}\times 11}{{{n}_{1}}+{{n}_{2}}}\] |
| or \[10.81\,{{n}_{1}}+10.81\,{{n}_{2}}=10\,{{n}_{1}}+11\,{{n}_{2}}\] |
| or \[0.81\,\,{{n}_{1}}=0.19\,{{n}_{2}}\] |
| or \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{0.19}{0.81}=\frac{19}{81}\] |
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