One mole of magnesium in the vapour state absorbed 1200 kJ of energy. If the first and second ionization energies of magnesium 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 of energy. If the first and second ionization energies of magnesium are 750 and 1450 kJ mol-1 respectively, the final composition of the mixture is (A) 31% Mg++ 69% Mg2+ (B) 69% Mg+ + 31% Mg2+ (C) 86% Mg+ + 14% Mg2+ (D) 14% Mg+ + 86% Mg2+

Tardigrade Admin · Accepted Answer

Energy absorbed in the ionisation of 1 mole of
Mg (g) to

M g^{+} 4 (g) = 750 k J

Energy left unused = 1200 - 750 = 450 kJ
% of

M g^{+}

(g) converted into

M g^{2 +}

(g)
=

\frac{450}{1450} \times 100

= 31%
. Thus the percentage of

M g^{+}

(g)
= 100 - 31 = 69%

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

$31$

$69$

$86$

$14$

Energy absorbed in the ionisation of 1 mole of
Mg (g) to $M g^{+} 4 (g) = 750 k J$
Energy left unused = 1200 - 750 = 450 kJ
% of $M g^{+}$ (g) converted into $M g^{2 +}$ (g)
= $\frac{450}{1450} \times 100$ = 31%
. Thus the percentage of $M g^{+}$ (g)
= 100 - 31 = 69%

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

31

69

86

14