First and second ionization energies of magnesium are 7.646 and 15.035 eV respectively. The amount of energy in kJ / mol needed to convert all the atoms of Magnesium into Mg 2+ ions present in 12 mg of magnesium vapours is: (Report your answer by multiplying with 10 and round it upto nearest integer) [Given: 1 eV =96.5 kJ mol -1 ]

Question

First and second ionization energies of magnesium are 7.646 and 15.035 eV respectively. The amount of energy in kJ / mol needed to convert all the atoms of Magnesium into Mg 2+ ions present in 12 mg of magnesium vapours is: (Report your answer by multiplying with 10 and round it upto nearest integer) [Given: 1 eV =96.5 kJ mol -1 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(12 mg /24 g )=0.5 × 10-3 moles Energy per atom =7.646+15.035=22.68 eV 22.68 × 96.48=21.88 × 102 kJ / mol E (needed) 21.88 × 102 × 0.5 × 10-3 10.94 × 10-1=1.094 kJ / mol

24g12mg​=0.5×10−3 moles Energy per atom =7.646+15.035=22.68eV 22.68×96.48=21.88×102kJ/mol E (needed) 21.88×102×0.5×10−3 10.94×10−1=1.094kJ/mol

$\frac{12 m g}{24 g} = 0.5 \times 1 0^{- 3}$ moles
Energy per atom $= 7.646 + 15.035 = 22.68 e V$
$22.68 \times 96.48 = 21.88 \times 1 0^{2} k J / m o l$
$E$ (needed) $21.88 \times 1 0^{2} \times 0.5 \times 1 0^{- 3}$
$10.94 \times 1 0^{- 1} = 1.094 k J / m o l$