Question

50kJ energy supplied to 1g Mg in its gaseous state. Find out the percentage of Mg+ [I.E. 1 = 740 kJ/ mole I.E. 2 = 1550 kJ/Mole]​

Answer 1

given, mass of Mg = 1g

we know, atomic mass of Mg = 24g/mol

so, no of moles of Mg = 1/24 = 0.0417 mol

so, energy required for first ionisation of 0.0417 mol of Mg , I.E = 0.0417 × 740 = 30.858 KJ

remaining energy is used for 2nd ionisation = 50KJ - 30.858 KJ = 19.172 KJ

so, let's find no of moles of Mg+ converted to Mg²+.

no of moles of Mg+ converted to Mg²+ = remaining energy/second ionisation energy

= 19.172/1550 = 0.0124 mol

so, Mg left as Mg+ = 0.0417 - 0.0124 = 0.0293 mol

now percentage of Mg+ = 0.0293/0.0417 × 100 = 70.26 %

percentage of Mg²+ = 0.0124/0.0417 × 100 = 29.74 %