An element X with relative atomic mass 16.2 contains two isotopes \(^{16}_{8}\)X with relative abundance of 90% and \(^{m}_{8}\)X with relative abundance of 10%. The value of m is
168X, m8X
[\(\frac{90}{100}\times 16\)] + [\(\frac{10}{100}\times m\)] = 16.2
14.4 + 0.1m = 16.2
0.1m = 1.8
m = \(\frac{1.8}{0.1}\)
m = 18
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