Chapter 16, Problem 69GP

To determine

To Find: Total charged of all electrons in an aluminum bar and net charge on the aluminum bar.

Answer to Problem 69GP

  $-6.96\times {10}^8\text{C}$ ; $0\text{C}$

Explanation of Solution

Given:

Mass of aluminum bar, $m$ = 15 kg

Molar mass of aluminum bar, $M$ = 27 kg

No. of electron per atom of Al, $n$ =13

Charge on an electron, $e=-1.6\times {10}^{-19}\text{C}$

Formula Used:

Total number of electrons:

  $N=\frac{m}{M}\times N_A\times n$

Here, m is the mass, M is the molar mass, n is the number of electrons per atom and $N_A$ is the Avogadro number

Calculation:

As 1 mole of Aluminum contains Avogadro number of atoms.

Where, $N_A=6.023\times {10}^{23}$ atoms.

The bar of aluminum contains: $N=\frac{m}{M}\times N_A\times n$

  $N=\left(\frac{15}{27\times {10}^{-3}}\right)\left(6.023\times {10}^{23}\right)(13)$

  $=4.35\times {10}^{27}$ electrons

Hence, total charge is $Q=Ne$

  $\begin{array}{l}Q=\left(-4.35\times {10}^{27}\right)\left(1.6\times {10}^{-19}\right)\text{C}\hfill \\ Q=-6.96\times {10}^8\text{C}\hfill \end{array}$

The net charge on the bar is $0\text{C}$ , as there are equal number of protons and electrons.