How much positive and negative charge is there in a cup of water?
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How much positive and negative charge is there in a cup of water?
Answer
\(1\mathrm{\ldotp }606\times {10}^{-7}{C}\)
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Solution
Let us assume that the mass of one cup of water \(=300\;\mathit{g.}\)
We know that the molecular mass of water is; \(M=18\;\text{g/mol}\) .
One mole of water \(=6.023\times 10^{23}\) molecules.
Therefore, one mole (which is\(6.023\times 10^{23}\) molecules) of water is\(18\;\mathit{g.}\)
i.e.,\(18\;g\)of water \(=6.023\times 10^{23}\;\text{molecules}\)
Therefore the number of molecules in one cup of water of \(300\;g\) can be found as,
\(300g\) of water\(=\frac{300}{18}\times 6.023\times 10^{23}\text{molecules}\)
Water \((H_2O)\) contains two hydrogen atoms and one oxygen atom.
i.e., \(10\) electrons and \(10\) protons.
Charge on one electron or proton is \(1.6\times 10^{-19}\mathit{C.}\)
Therefore the charge present in a cup of water of mass \(300\;g\) equals to,
\((\frac{300}{18})\times 6\mathrm{\ldotp }023\times {10}^{23}\times 10\times 1\mathrm{\ldotp }6\times {10}^{−19}C=1\mathrm{\ldotp }606\times {10}^{-7}{C}\)
As water is neutral, the amount of negative charge will be equal to the amount of positive charge which is equivalent to \(1\mathrm{\ldotp }606\times {10}^{-7}{C}\).
We know that the molecular mass of water is; \(M=18\;\text{g/mol}\) .
One mole of water \(=6.023\times 10^{23}\) molecules.
Therefore, one mole (which is\(6.023\times 10^{23}\) molecules) of water is\(18\;\mathit{g.}\)
i.e.,\(18\;g\)of water \(=6.023\times 10^{23}\;\text{molecules}\)
Therefore the number of molecules in one cup of water of \(300\;g\) can be found as,
\(300g\) of water\(=\frac{300}{18}\times 6.023\times 10^{23}\text{molecules}\)
Water \((H_2O)\) contains two hydrogen atoms and one oxygen atom.
i.e., \(10\) electrons and \(10\) protons.
Charge on one electron or proton is \(1.6\times 10^{-19}\mathit{C.}\)
Therefore the charge present in a cup of water of mass \(300\;g\) equals to,
\((\frac{300}{18})\times 6\mathrm{\ldotp }023\times {10}^{23}\times 10\times 1\mathrm{\ldotp }6\times {10}^{−19}C=1\mathrm{\ldotp }606\times {10}^{-7}{C}\)
As water is neutral, the amount of negative charge will be equal to the amount of positive charge which is equivalent to \(1\mathrm{\ldotp }606\times {10}^{-7}{C}\).
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