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Molecular Mass

The I. II and...

Question

The I. II and III I.E. of al are 578, 1817 and 2745 kJ/mol $^{- 1}$ respectively. Calculate the energy required to convert all the atoms of Al to Al $^{3 -}$ present in 270 mg of Al vapours.

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Solution

Energy required to convert 1 mole of $A l$ to $A l^{3 +}$ = $I E_{1} + I E_{2} + I E_{3} = 578 + 1817 + 2745 = 5140 k J$
$∴$ Moles of $A l$ in 270 mg = $\frac{270 \times 10^{- 3}}{27} = 0.01 m o l e s$
Therefore, total energy required = $5140 \times 0.01 k J = 51.4 k J$

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