Question

A cup contains 250g of water. find number of positive charges present in the cup of water?

Answer 1

A cup of water contains 250 gram of water. As the density of pure water is unity (in CGS unit system) therefore, the number of moles of water in 250 gram of water will be,
$$\begin{gathered} n=\frac{250gm}{molarwt.ofwater\left(H_{2}O\right)} \\ n=\frac{250gm}{2\times 1+1\times 16} \\ n=\frac{250gm}{18} \\ n=13.88moles \end{gathered}$$

As we know that,
$1mole=6.023\times {10}^{23}numberof{H}_{2}Omolecules$
Therefore,
250 g water will have $13.88\times 6.023\times {10}^{23}numberofmoleculesof{H}_{2}O=8.36\times {10}^{24}numberofmoleculesofwater$
Since, one water molecule contains 2 hydrogen atoms (that is 2 protons) and an oxygen atom (that is 8 protons).
Therefore,
$$\begin{gathered} Totalnumberofprotonsinwatermoleculeis=10 \\ Totalnumberofelectrons=10(aswaterisnuetralinnature) \end{gathered}$$

So 250 g water will consists of $10\times 8.36\times {10}^{24}=8.36\times {10}^{25}numberofprotons$.
As we know that the charge on proton is,
$p=1.6\times {10}^{-19}C$
Therefore, the total positive charge of 250 g water will be,
$$\begin{gathered} q=1.6\times {10}^{-19}\times 8.36\times {10}^{25}C \\ q=13.37\times {10}^{6}C \\ q=13.3MC \end{gathered}$$
So a cup of 250 gms of water consists of 13.3 Mega coulomb of positive and negative charge, so that the overall charge is zero.