Question

Estimate the total negative charge in $1g$ of water. Given that Avogadro number $=6.02\times 10^{23}$ and molecular weight of water $=18$

• A. $6.25\times 10^{4}C$
• B. $5.35\times 10^{4}C$
• C. $1.76\times 10^{4}C$
• D. $1.25\times 10^{4}C$

Answer 1

Answer: (B)

One molecule of water $\left(H_{2}O\right)$ contains $2+8$ i.e., 10 electrons. Number of molecules in $1g$ 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 $1g$ of water,
$n^{\prime }=n\times 10=3.344\times 10^{22}\times 10=3.344\times 10^{23}$
The negative charge possessed by $1g$ of water,
$q=n^{\prime }e=3.344\times 10^{23}\times 1.6\times 10^{-19}=5.35\times 10^{4}C$