Q. $I^{st}\ and\ II^{nd}\ IE$ of Mg are 7.646 and 15.035 eV respectively. The amount of energy needed to convert all the atoms of magnesium into $Mg^{2 +}$ ions present in 12 mg of magnesium vapours is [Given, 1eV = 96.5 kJ $mol^{- 1}$]
A) 1.5
B) 2.0
C) 1.1
D) 0.5
Explanation
(c)
Total energy required for the conversion of one Mg atom into $Mg^{2 +}is = IE_{1} + IE_{2}$
$= 7.646 + 15.035\ eV$
$= 22.681\ eV$
$= 2188.6\ kJ\ mol^{- 1}$
Moles of $Mg = \frac{12 \times 10^{- 3}}{24}$
$= 0.5 \times 10^{- 3}$
$\therefore$ The energy required to convert $0.5 \times 10^{- 3}$ mol Mg into
$Mg^{2 +} = 0.5 \times 10^{- 3} \times 2188.6$
$= 1.09 \approx 1.1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.