Question

How much positive and negative charge is there in a cup of water ?
Take mass of 1 cup of water as 250 g. Mass of $6.02\times {10}^{23}$ molecules of water is 18 g.

Answer 1

A water molecule $\left(H_{2}O\right)$ contain no. of proton $\left(np\right)=8+2\times 1=10$ and so no. of electrons $=20$

Electroneutrating principle statter state

net positive charge $=$ net negative charge

As $L$ cup of water holder $250g$ water,

no. of moles $=\frac{givenmass}{molarmass}=\frac{250}{18}=\frac{125}{9}moles$

$1$ mol of water contains $N_A$ no. of molecular

$\Rightarrow$ Total no. of molecular $=\frac{125}{9}\times 6.02\times {10}^{24}$

$=8.361\times {10}^{24}$

so, net positive charge present $=20\times 8.361\times {10}^{24}\times 1.6\times {10}^{-19}$ (charge of photon)

$\Rightarrow Q=2.68\times {10}^{7}C$

Thus, net negative charges presents $=-Q$

$=-2.68\times {10}^{7}C$