One mole of magnesium in the vapour state absorbed 1200 kJ mol -1 of energy. If the first and second ionization 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 of energy. If the first and second ionization 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) 36 % Mg ++64 % Mg 2+ (C) 69 % Mg ++31 % Mg 2+ (D) 31 % Mg ++69 % Mg 2+

Tardigrade Admin · Accepted Answer

Number of moles of one g of Mg =(1/24) =0.0417 1 g of Mg ( g ) absorbs =(1200/24)= 50 kJ Energy required to convert Mg ( g ) to Mg +( g ) =0.0417 × 750=31.275 kJ Remaining energy =50-31.275 =18.725 kJ Number of moles of Mg 2+ formed =(18.725/1450) =0.013 Thus remaining Mg +will be =0.0417-0.013=0.0287 So, % Mg 2+ =(0.0287/0.0417) × 100=68.82 % % Mg 2+ =100-68.82=31.18 %

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

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

$36% M g^{+} + 64% M g^{2 +}$

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

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

Q. One mole of magnesium in the vapour state absorbed 1200kJmol1 of energy. If the first and second ionization energies of Mg are 750 and 1450kJmol−1 respectively, the final composition of the mixture is

86%Mg++14%Mg2+

36%Mg++64%Mg2+

69%Mg++31%Mg2+

31%Mg++69%Mg2+

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

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

$36% M g^{+} + 64% M g^{2 +}$

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

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