A) 201 amu
B) 202 amu
C) 199 amu
D) 200 amu
Correct Answer: D
Solution :
Average isotopic mass of \[X=\frac{200\times 90+199\times 8+202\times 2}{90+8+2}\] \[=\frac{18000+1592+404}{100}=\frac{19996}{100}\] \[=199.96\,\text{amu}\]You need to login to perform this action.
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