Question

How many atoms are in a drop of mercury that has a diameter of 1.0 mm? (volume of a sphere is 4πr3/3; density of mercury = 13.6 g/cm3)?

Answer 1

The diameter of drop of mercury is 1.0 mm. The volume of drop is equal to the volume of sphere, thus, volume of mercury drop will be:

$V=\frac{4}{3}\pi r^{3}$

Here, r is radius of drop.

Radius of drop can be calculated as:

$r=\frac{d}{2}=\frac{1 mm}{2}=0.5 mm$

Convert the unit into cm

1 mm=0.1 cm

thus,

0.5 mm=0.05 cm

Volume of sphere will be:

$V=\frac{4}{3}(3.14) (0.05 cm)^{3}=5.23\times 10^{-4}cm^{3}$

Now, density of mercury is $13.6 g/cm^{3}$

Mass can be calculated as follows:

$m=d\times V=13.6 g/cm^{3}\times 5.23\times 10^{-4}cm^{3}=7.12\times 10^{-3}g$

Molar mass of mercury is 200.59 g/mol thus, number of moles will be:

$n=\frac{m}{M}=\frac{7.12\times 10^{-3}g}{200.59 g/mol}=3.55\times 10^{-5} mol$

In 1 mole of a substance there are $6.023\times 10^{23}$ atoms thus, $3.55\times 10^{-5} mol$ will have:

$N=3.55\times 10^{-5}mol\times 6.023\times 10^{23}atoms/mol=2.14\times 10^{19} atoms$

Therefore, number of atoms in a drop of mercury will be $2.14\times 10^{19} atoms$.