One mole of magnesium in the vapour state absorbed 1200 kJ mol -1 energy. If the first and second ionisation energies of Mg are 750 and 1450 kJ mol -1 respectively, the final composition of the mixture is

Question

One mole of magnesium in the vapour state absorbed 1200 kJ mol -1 energy. If the first and second ionisation energies of Mg are 750 and 1450 kJ mol -1 respectively, the final composition of the mixture is (A) 86 % Mg ++14 % Mg 2+ (B) 69 % Mg ++31 % Mg 2+ (C) 14 % Mg ++86 % Mg 2+ (D) 31 % Mg ++69 % Mg 2+

Tardigrade Admin · Accepted Answer

Energy absorbed in the ionisation of 1 mole of Mg to Mg +(g)=750 kJ Energy left unused =1200-750=450 kJ % of Mg +(g) converted into Mg 2+(g) =(450/1450) × 100=31 % Hence, the % of Mg +(g)=100-31=69 %

Q. One mole of magnesium in the vapour state absorbed $1200 k J m o l^{- 1}$ energy. If the first and second ionisation energies of $M g$ are $750$ and $1450 k J m o l^{- 1}$ respectively, the final composition of the mixture is

$86% M g^{+} + 14% M g^{2 +}$

$69% M g^{+} + 31% M g^{2 +}$

$14% M g^{+} + 86% M g^{2 +}$

$31% M g^{+} + 69% M g^{2 +}$

Energy absorbed in the ionisation of $1$ mole of $M g$ to $M g^{+} (g) = 750 k J$
Energy left unused $= 1200 - 750 = 450 k J$ $%$ of $M g^{+} (g)$ converted into $M g^{2 +} (g)$
$= \frac{450}{1450} \times 100 = 31%$
Hence, the $%$ of $M g^{+} (g) = 100 - 31 = 69%$

Q. One mole of magnesium in the vapour state absorbed 1200kJmol−1 energy. If the first and second ionisation energies of Mg are 750 and 1450kJmol−1 respectively, the final composition of the mixture is

86%Mg++14%Mg2+

69%Mg++31%Mg2+

14%Mg++86%Mg2+

31%Mg++69%Mg2+