Chapter 4: Q4E (page 713)

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?

Expert verified

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

Step by step solution

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

So, the number of lead atoms is,

$N = {nN}_{A}$

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,

$\begin{aligned} N = (6.022 \times 10^{23} atoms / mol) (\frac{10.8 g}{197 g / mol}) \\ = 3.300 \times 10^{22} atoms \end{aligned}$

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

$\begin{aligned} n_{p} = (79 protons/atom) (3.300 \times 10^{22} atoms) \\ = 2.61 \times 10^{24} protons. \end{aligned}$

Thus,Charge of the protons is,

$\begin{aligned} Q = n_{p} (1.60 \times 10^{- 19} C / proton) \\ = 4.18 \times 10^{5} C \end{aligned}$

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

$\begin{aligned} ne = n p \\ = 2.61 \times 10^{24} electrons \end{aligned}$

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

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