Question

The energy required to convert all atoms present in 1.2 g magnesium to Mg+2 ions if iE1 and IE2 of magnesium are 120kJ per mole and 240 Kj per mole respectively?

Answer 1

This is the correct answer
Mark as brainliest

Answer 2

Total amount of energy needed to convert Mg (g) into $\bold{\mathrm{Mg}^{2+}}$ ion is $\bold{18 \ k J.m o l^{-1}}$

Explanation:

The conversion of Mg (g) into $\mathrm{Mg}^{2+}$ is as follows:

$M g(g)+I E_{1} \rightarrow M g^{+}(g)+e^{-} ; \quad I E_{1}=120 \ k J m o l^{-1}$

$M g^{+}(g)+I E_{2} \rightarrow M g^{2+}(g)+e^{-} ; \quad I E_{2}=240 \ k J m o l^{-1}$

Total amount of energy needed to convert Mg (g) into $\mathrm{Mg}^{2+}$ ion.

$\mathrm{Mg}^{2+} \mathrm{ion}=\mathrm{I} \mathrm{E}_{1}+\mathrm{I} \mathrm{E}_{2}$

$=120+240$

$\mathrm{Mg}^{2+} \text { ion }=360 \ \mathrm{kJ} \mathrm{mol}^{-1}$

$\text { Molecular mass of magnesium }=24 \ \mathrm{gm} / \mathrm{mole}$

$\text { One mole of magensium }=360 \ \mathrm{kJ} \mathrm{mol}^{-1}$

$1.2 \ \mathrm{g}=\frac{1.2 \times 360}{24}=18 \ \mathrm{kJ} \mathrm{mol}^{-1}$