Question

The ${△}_i{H}_1$ and ${△}_i{H}_2$of Mg(g) are 740 and 1450 kJ ${△}_i{H}_2$ ${mol}^{-1}$. Calculate the percentage of ${Mg}^{+}\left(g\right)$ and ${Mg}^{2+}\left(g\right)$ if 1 g of Mg(g) absorbs 50 kJ of energy.

• A. 75% and 25%
• B. 66.67% and 33.33%
• C. 60% and 40%
• D. None of thse

Answer 1

The correct option is B 66.67% and 33.33%

$(Mg\to M{g}^{+}+e)740kJ/mol$

$\frac{1}{24}\to M{g}^{+}\to M{g}^{2+}+e^{-}1450kJ/mol$

$\perp 9Mg\to \frac{1}{24}$ Moles

Assume is the ratio of moles of

$M\to M{g}^{+}$ $M{g}^{+}$ & $M{g}^{2+}$

$n\to M{g}^{2+}$

for $Mg\to \frac{1}{24}$ moles $\to \frac{1}{24}\times 740\Rightarrow 30.8kJ$

Rest $=50-30.8=19.2kJ/mol$

$\to M{g}^{2+}\to 1450\to \perp$

$19.2\to \frac{1}{1450}\times 19.2$

$=0.013moles$

So, $M{g}^{+}=\frac{1}{24}$ Mole, $M{S}^{2+}\to 0.013$ moles

$-0.013$

$\downarrow$

$0.028$ $\%M{g}^{+}\to \frac{0.028}{0.042}$

$M{g}^{+}\Rightarrow \mathrm{66.67.1}$

$M{g}^{2+}$

$\Rightarrow \frac{0.013}{0.042}$

$\Rightarrow 33.331$