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An ion with mass number 37 possesses one unit of negative charge. If the ion contains \(11.1 \%\) more neutrons than the electrons, find the symbol of the ion.
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Suppose number of electrons in the ion \(=x\)
Then number of neutrons
\(=\mathrm{x}+\frac{11.1}{100} \mathrm{x}=1.111 \mathrm{x}\)
No. of electrons in the neutral atom \(=x-1\)
\(\therefore\) No. of protons \(=x-1\)
Mass number \(=\) No. of neutrons \(+\) No. of protons \(\therefore \quad 37=1.111 \mathrm{x}+\mathrm{x}-1\) or
\(2.111 \mathrm{x}=38\) or \(\mathrm{x}=18\)
Hence, the symbol of the ion will be \({ }_{17}^{37} \mathrm{Cl}^{-1}\)
Then number of neutrons
\(=\mathrm{x}+\frac{11.1}{100} \mathrm{x}=1.111 \mathrm{x}\)
No. of electrons in the neutral atom \(=x-1\)
\(\therefore\) No. of protons \(=x-1\)
Mass number \(=\) No. of neutrons \(+\) No. of protons \(\therefore \quad 37=1.111 \mathrm{x}+\mathrm{x}-1\) or
\(2.111 \mathrm{x}=38\) or \(\mathrm{x}=18\)
Hence, the symbol of the ion will be \({ }_{17}^{37} \mathrm{Cl}^{-1}\)
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