Question

How many electrons are in 0.990 oz of a pure gold coin?

Answer 1

Atomic number of gold, $Au$ = 79 means there are 79 electrons in per atom of gold.

Atomic mass of gold = 196.967 gm

0.990 oz of gold is given.

Now, converting $oz to gm$

$1 oz = 28.3495 gm$

So, $0.990 oz = 0.990\times 28.3495 gm = 28.066 gm$

Now, for determining the number of moles, the formula for number of moles is:

$number of moles = \frac{given mass}{Molar mass}$

Substituting the values:

$number of moles = \frac{28.066}{196.967} = 0.1425 mole$

1 mole of $Au = 6.022\times 10^{23} atoms of Au$

So, for 0.1425 mole of $Au = 6.022\times 10^{23}\times 0.1425 atoms of Au = 0.8581 \times 10^{23} atoms of Au$

Since 1 atom of gold has 79 electrons so,

Number of electrons = $0.8581 \times 10^{23}\times 79 = 67.7899\times 10^{23} electrons$

So, number of electrons in 0.990 oz of pure gold coin is $67.7899\times 10^{23}$.