Answer
(a) The number of electrons are in one liter of water $\approx3.35\times10^{26}$
(b) The net charge of all the electrons contained in one liter water $=-5.36\times10^{7} C$
Work Step by Step
(a) $1$ liter water $=$ $1000$ $g$ water
Mass of $1$ mole water $=$ $18$ $g$
Therefore, one liter water $=\frac{1000}{18}$ moles water
$1$ mole water contains $6.022\times10^{23}$ (avogadro's number) number of water molecules.
Therefore number of molecules in $1$ liter water $=\frac{1000}{18}\times6.022\times10^{23}$
Given, each water molecule has 10 electrons
Therefore, the number of electrons are in one liter of water $=\frac{1000}{18}\times6.022\times10^{23}\times10$
$\approx3.35\times10^{26}$
(b) Charge of an electron $-1.60\times10^{-19} C$
Therefore the net charge of all the electrons contained in one liter water
$=-(3.35\times10^{26}\times1.60\times10^{-19}) C$
$=-5.36\times10^{7} C$
