1g of Mg atoms in the vapour phase absorbs 50.0kJ of energy. Find the composition of Mg2+. formed as a result of absorption of energy.IE1 and IE2 for Mg are 740 and 1450kJ mol?1 respectively.

Answer:

A mole of Mg = 1/24

These many moles of Mg is converted to Mg+ and Mg2+. The energy consumed for Mg to Mg+ and Mg2+ conversation is

740 + 1450 =2190.

Let us consider x grams of Mg+ ions and y grams of Mg2+ ions.

x + y =1 (Total mass in 1 gram)

Moles of Mg+ ions = x/24

Moles of Mg2+ ions = y/24

Energy absorbed for generating Mg+ ions = x/24 × 740

Energy absorbed for forming Mg2+ ions= y/24 × 2190

Total energy absorbed (E) = x/24 × 740 + y/24 × 2190

50 = x/24 × 740 + y/24 × 2190

1200 = 740x + 2190y

120 = 74x + 219y

since x + y =1

x = 1 – y

120 = 74(1 – y) + 219y

120 = 74 – 74y + 219y

120 = 145y + 74

46 = 145y

y = 0.3172

The mass of Mg2+ ions is 0.3172g

% of Mg2+ ions = 0.31721 × 100 = 31.72

Also, % of Mg+ ions =100 – 31.72 = 68.28.

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