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How much positive and negative charge is there in a cup of water?
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## QuestionPhysicsClass 12

How much positive and negative charge is there in a cup of water?

$$1\mathrm{\ldotp }606\times {10}^{-7}{C}$$
4.6
4.6

## 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 $$1\mathrm{\ldotp }606\times {10}^{-7}{C}$$.