Chapter 16, Problem 9P

To determine

The total charge of all electrons in given amount of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ .

Answer to Problem 9P

The total charge of all electrons in 1 kilogram of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ is $-5.34\times {10}^7\text{C}$ .

Explanation of Solution

Given:

Net mass, $m=1\text{kg}$

Formula used:

Apply molar mass formula to calculate the number of molecules.

Calculation:

Since, an oxygen atom has 8 electrons and a hydrogen atom has 1 electron. So, there are total 10 electrons in water molecule $\left({\text{H}}_{\text{2}}\text{O}\right)$ . Let the molar mass of water is $18.02$ kilograms per mole.

Calculate the number of molecules in 1 kilogram of water.

  $\begin{array}{c}\text{Number}\text{of}\text{molecules}=1\text{kg}\times \frac{1\text{mole}}{0.018\text{kg}}\times \frac{6.023\times {10}^{23}}{1\text{mole}}\\ =3.34\times {10}^{25}\end{array}$

Calculate the number of electrons.

  $\begin{array}{c}\text{Number}\text{of}\text{electrons}=3.34\times {10}^{25}\times 10\\ =3.34\times {10}^{26}\text{electrons}\end{array}$

Calculate the total charge.

  $\begin{array}{c}\text{Total charge}=3.34\times {10}^{26}\times \left(-1.6\times {10}^{-19}\right)\text{C}\\ \text{=}-5.34\times {10}^7\text{C}\end{array}$

Conclusion:

Therefore, the total charge of all electrons in 1 kilogram of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ is $-5.34\times {10}^7\text{C}$ .