What is the total charge of all the electrons in 1.0 L of liquid water?

A water molecule has no overall charge, but each hydrogen atom has a little positive charge and the oxygen atom has a slight negative charge.

The charge is represented by $Q$and for an object its charge equals the number of carriers N times the charge of carrier which is $-e$for electrons and $+e$for protons. Therefore,

$Q=Ne$

We are given,

$e=-1.6\times {10}^{-19}C$

$p=1kg/L$where $p$is the density of water,

$V=1L$where $V$is the volume of water.

The mass represents the product of volume and the density. Therefore mass of water is

$$\begin{gathered} m=pV \\ m=(1kg/L)(1L) \\ m=1kgor1000g \end{gathered}$$

The number of moles is calculated by dividing the mass of the object (m) with its molecular weight (M). We are given here, $M=18g/mole$where M represents molecular weight.

The number of moles is equal to,

$$\begin{gathered} n=\frac{m}{M} \\ n=\frac{1000g}{18g/mol} \\ n=55.55moles \end{gathered}$$

According to Avogadro's number $1moles=6.02\times {10}^{23}molecule$

The total number of electrons in $55.55moles$ow water is calculated as:

$$\begin{gathered} N=10electrons/molecule\left(n{N}_A\right) \\ N=10electrons/molecule(55.55oles)(6.02\times {10}^{23}molecules) \\ N=3.34\times {10}^{26}electrons \end{gathered}$$

Substitute the values in the equation localid="1648551824585" $Q=Ne$

$$\begin{gathered} Q=Ne \\ Q=(3.34\times {10}^{26}electrons)(-1.6\times {10}^{-19}C) \\ Q=-5.34\times {10}^{7}C \end{gathered}$$