Thank you for reporting, we will resolve it shortly
Q.
Estimate the total negative charge in $1 g$ of water. Given that Avogadro number $=6.02 \times 10^{23}$ and molecular weight of water $=18$
Electric Charges and Fields
Solution:
One molecule of water $\left( H _{2} O \right)$ contains $2+8$ i.e., 10 electrons. Number of molecules in $1 g$ of water,
$n=\frac{\text { Avogadro number }}{\text { atomic weight }}=\frac{6.02 \times 10^{23}}{18}=3.344 \times 10^{22}$
Therefore, number of electrons in $1 g$ of water,
$n^{\prime}=n \times 10=3.344 \times 10^{22} \times 10=3.344 \times 10^{23}$
The negative charge possessed by $1 g$ of water,
$q=n^{\prime} e=3.344 \times 10^{23} \times 1.6 \times 10^{-19}=5.35 \times 10^{4} C$