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1g of Mg atoms in the vapour phase absorbs50.0kJ of energy. Find the composition of Mg+2 formed as a result of absorption of energy. IE1 andIE2 for Mg are 740 and 1450 kJmol-1 respectively?


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Solution

Step 1: Given data

Weight of Mgatoms present = 1g

Energy absorbed by 1g of Mg atoms in the vapor phase = 50.0kJ

First ionization energy for Mg(IE1) = 740kJmol-1

Second ionization energy for Mg(IE2) = 1450kJmol-1

Step 2: Formula used.

The number of moles is calculated using the formula: No.ofmoles=GivenmassMolarmass

The energy absorbed is calculated using the formula: Energyabsorbed=IE1+IE2

The energy consumed for Mg to Mg+ and Mg+2 conversion is: 740+1450=2190kJmol-1

Step 3: Calculation of the number of moles.

The number of moles of Mg atoms is calculated using the formula: No.ofmoles=124

Hence 124 moles of Mg is converted to Mg+ and Mg+2.

Let us consider xgrams of Mg+ ions and y grams of Mg+2 ions.

x+y=1....(1) as total mass is 1g

MolesofMg+ions=x24

MolesofMg+2ions=y24

Step 4: Calculation of energy absorbed.

The energy absorbed for generating x24moles of Mg+ ions =x24×740kJmol-1

The energy absorbed for generating y24moles of Mg+2 ions= y24×2190kJmol-1

Total energy absorbed =x24×740+y24×2190kJmol-1

50=x24×740+y24×21901200=740x+2190y120=74x+219y....(2)

Step 5: Calculation of percentage of Mg+2ions.

From equation (1) ,

x+y=1

x=1y

Putting value of x in equation 2:

120=74(1y)+219y120=7474y+219y120=145y+7446=145yy=0.3172

Hence, the mass of Mg+2 ions is 0.3172g

% of Mg+2 ions = 0.31721×100=31.72

Hence, 31.72% of Mg+2 ions is present.


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