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How much positive and negative charge is there in a cup of water $(250\;gm)$?
$7.58 \times 10^{9} \;C$
$3.65 \times 10^{6} \;C$
$1.34 \times 10^{7} \;C$
$2.68 \times 10^{8} \;C$
Solution
Let us assume that the mass of one cup of water is $250 \,g .$ The molecular mass of water is $18\, g$.
Thus, one mole $\left( =6.02 \times 10^{23} \text { molecules }\right)$ Therefore the number of molecules in one cup of water is $(250 / 18) \times 6.02 \times 10^{23}$
Each molecule of water contains two hydrogen atoms and one oxygen atom. i.e… $10$ electrons and $10$ protons. Hence the total positive and total negative charge has the same magnitude. It is equal to $(250 / 18) \times 6.02 \times 10^{23} \times 10 \times 1.6 \times 10^{-19}\, C$$ =1.34 \times 10^{7} \;C$
