The first and second ionization potentials of an element M (atomic weight = 25) are 800 and 1500 kJ/mol respectively. Calculate the percentage of ions formed if 5 (g) of M(g) absorbs 250 kJ of energy.

Question

The first and second ionization potentials of an element M (atomic weight = 25) are 800 and 1500 kJ/mol respectively. Calculate the percentage of $M_{(g)}^{2 +}$ ions formed if 5 (g) of M(g) absorbs 250 kJ of energy.

Smart Achievers · Accepted Answer

(30)

$M ⟶ M^{+} + e^{-}$ $Number of moles = \frac{5}{25} = 0.2$

$Energy required to form M^{+} ions = 0.2 \times 800 = 160 {kJ mol}^{- 1}$

$Remaining energy = 90 kJ$

$This is used to convert M^{+} to M^{2 +}$

$Number of moles of M^{2 +} formed = \frac{90}{1500} = 0.06$

$% M^{2 +} = \frac{0.06}{0.2} \times 100 = 30 %$

Solution

Q.
The first and second ionization potentials of an element M (atomic weight = 25) are 800 and 1500 kJ/mol respectively. Calculate the percentage of $M_{(g)}^{2 +}$ ions formed if 5 (g) of M(g) absorbs 250 kJ of energy.

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Solution

Q. The first and second ionization potentials of an element M (atomic weight = 25) are 800 and 1500 kJ/mol respectively. Calculate the percentage of M(g)2+ ions formed if 5 (g) of M(g) absorbs 250 kJ of energy.

Ans. (30) M⟶M++e- Number of moles=525=0.2 Energy required to form M+ ions=0.2×800=160 kJ mol-1 Remaining energy=90 kJ This is used to convert M+ to M2+ Number of moles of M2+ formed=901500=0.06 %M2+=0.060.2×100=30%

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Q.
The first and second ionization potentials of an element M (atomic weight = 25) are 800 and 1500 kJ/mol respectively. Calculate the percentage of $M_{(g)}^{2 +}$ ions formed if 5 (g) of M(g) absorbs 250 kJ of energy.