Question

You have a pure (24-karat) gold ring of mass 10.8 g. Gold has an atomic mass of 197 g/mol and an atomic number of 79. (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?

Short answer

1. The total number and charge of protons in the ring are $2.61\times {10}^{24}4.18\times {10}^{5}C$respectively.
2. The number of electrons in the ring when there is no charge is$2.61\times {10}^{24}\mathrm{electron}s$.

Step-by-step solution

(a) Determination of the total number and charge of protons in the ring.

The number atoms in 1 mole of a substance are the Avogadro number N_A.

So, the number of lead atoms is,

$\mathrm{N}={\mathrm{nN}}_{\mathrm{A}}$

Here, n is the number of moles and N_A is the Avogadro Number.

Now,the atomic mass of lead is,

M = 197g/mol

So the number of atoms in a mass of m = 10.8 is,

$\begin{array}{r}\mathrm{N}=\left(6.022\times {10}^{23}\text{atoms}/\mathrm{mol}\right)\left(\frac{10.8\mathrm{g}}{197\mathrm{g}/\mathrm{mol}}\right)\\ =3.300\times {10}^{22}\text{atoms}\end{array}$

Atomic number is 79, thus the number of protons is,

$\begin{array}{r}{n}_p=\left(79\text{protons/atom}\right)\left(3.300\times {10}^{22}\text{atoms}\right)\\ =2.61\times {10}^{24}\text{protons.}\end{array}$

Thus,Charge of the protons is,

$\begin{array}{r}Q=n_p\left(1.60\times {10}^{-19}\mathrm{C}/\text{proton}\right)\\ =4.18\times {10}^5\mathrm{C}\end{array}$

(b) Determination of the number of electrons in the ring when there is no charge.

The number of electrons is always equal to the number of protons. So,

$\begin{array}{r}\mathrm{ne}=\mathrm{n}\mathrm{p}\\ =2.61\times {10}^{24}\text{electrons}\end{array}$

Thus, the number of electron son the ring is $2.61\times {10}^{24}$electrons .