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

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,
$$\begin{gathered} n=\frac{250 gm}{molar wt. of water\left(H_{2}O\right)} \\ n=\frac{250 gm}{2\times 1+1\times 16} \\ n=\frac{250 gm}{18} \\ n=13.88 moles \end{gathered}$$

As we know that,
$1 mole=6.023\times {10}^{23} number of H_{2}O molecules$
Therefore,
 250 g water will have $13.88\times 6.023\times {10}^{23} number of molecules of H_{2}O=8.36\times {10}^{24} number of molecules of water$
Since, one water molecule contains 2 hydrogen atoms (that is 2 protons) and an oxygen atom (that is 8 protons).
Therefore,
$$\begin{gathered} Total number of protons in water molecule is=10 \\ Total number of electrons=10 (as water is nuetral in nature) \end{gathered}$$
 
So 250 g water will consists of $10\times 8.36\times {10}^{24}=8.36\times {10}^{25} number of protons$.
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,
 $$\begin{gathered} q=1.6\times {10}^{-19}\times 8.36\times {10}^{25} C \\ q=13.37\times {10}^6 C \\ q=13.3 MC \end{gathered}$$
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.