Question

Given R=1.0974 x 107 m-1 and h=6.626 x 10–34 Js. The
ionization energy of one mole of Li+2 ions will be as follows-​

Answer 1

Explanation:

We have been given some terms which are given below.

• R = 1.0974 * 10^7 m
• h = 6.626 * 10^-34 j/s

We have to find the ionization energy of one mole of Li^2+.

• General expression of ionization energy = RZ²hc

For Li^2+, We have:

• Z(No. of Protons) = 3

Put the value of given values:

Ionization energy = RZ²hc

= (1.0974 * 10^7)*(3)²*(6.626 * 10^-34)(3*10^8)

= 1.96 * 10^-17 j

For one of mole of Li^2+,

IE = Avogardo constant * IE

IE = 6.022*10^23 mol^-1* 1.96 * 10^-17 j

IE = 1.118 * 10^7 j/mol

Or, IE = 11180 kg/mol

• Therefore, the required ionization energy of one mole of Li+2 ions will be 11180 kg/mol.