Question

The ionization energy of gold is 8.90x10^2 kJ/mol. How much energy, in kilojoules, is required to ionize 79 atoms of gold?

Answer 1

ANSWER:

Given,

Ionization energy of gold = 8.90 x 10² kJ/mol

Number of atoms of gold = 79 atoms

Now,

1 mole = 6.022 x 10²³ atoms

1 atom = 1/(6.022 x 10²³) mole

1 atom = 1.66 x 10^-24 mole

And,

79 atoms of gold = 79 x 1.66 x 10^-24 mole

= 1.31 x 10^-22 mole

Here,

ionization energy of 1 mole of gold = 8.90 x 10² kJ/mol

So, ionization energy of 1.31 x 10^-22 mole of gold = 1.31 x 10^-22 mol x 8.90 x 10² kJ/mol

= 1.17 x 10^-19 kJ

Hence, 1.17 x 10^-19 kJ energy is required to ionize 79 atoms of gold.