Question

# $1\mathrm{g}$ of $\mathrm{Mg}$ atoms in the vapour phase absorbs$50.0\mathrm{kJ}$ of energy. Find the composition of ${\mathrm{Mg}}^{+2}$ formed as a result of absorption of energy. ${\mathrm{IE}}_{1}$ and${\mathrm{IE}}_{2}$ for $\mathrm{Mg}$ are $740$ and $1450$ ${\mathrm{kJmol}}^{-1}$ respectively?

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Solution

## Step 1: Given data Weight of $\mathrm{Mg}$atoms present = $1\mathrm{g}$Energy absorbed by $1\mathrm{g}$ of $\mathrm{Mg}$ atoms in the vapor phase = $50.0\mathrm{kJ}$First ionization energy for $\mathrm{Mg}$(${\mathrm{IE}}_{1}$) = $740{\mathrm{kJmol}}^{-1}$Second ionization energy for $\mathrm{Mg}$(${\mathrm{IE}}_{2}$) = $1450{\mathrm{kJmol}}^{-1}$Step 2: Formula used. The number of moles is calculated using the formula: $No.ofmoles=\frac{Givenmass}{Molarmass}$The energy absorbed is calculated using the formula: $Energyabsorbed=I{E}_{1}+I{E}_{2}$The energy consumed for $\mathrm{Mg}$ to ${\mathrm{Mg}}^{+}$ and ${\mathrm{Mg}}^{+2}$ conversion is: $740+1450=2190kJmo{l}^{-1}$Step 3: Calculation of the number of moles. The number of moles of $\mathrm{Mg}$ atoms is calculated using the formula: $No.ofmoles=\frac{1}{24}$Hence $\frac{1}{24}$ moles of $\mathrm{Mg}$ is converted to ${\mathrm{Mg}}^{+}$ and ${\mathrm{Mg}}^{+2}$.Let us consider $x$grams of ${\mathrm{Mg}}^{+}$ ions and $y$ grams of ${\mathrm{Mg}}^{+2}$ ions.$⇒x+y=1....\left(1\right)$ as total mass is 1g$⇒MolesofM{g}^{+}ions=\frac{x}{24}$$⇒MolesofM{g}^{+2}ions=\frac{y}{24}$Step 4: Calculation of energy absorbed. The energy absorbed for generating $\frac{x}{24}$moles of ${\mathrm{Mg}}^{+}$ ions =$\frac{x}{24}×740kJmo{l}^{-1}$The energy absorbed for generating $\frac{y}{24}$moles of ${\mathrm{Mg}}^{+2}$ ions= $\frac{y}{24}×2190kJmo{l}^{-1}$Total energy absorbed =$\left(\frac{x}{24}×740+\frac{y}{24}×2190\right)kJmo{l}^{-1}$$⇒50=\frac{x}{24}×740+\frac{y}{24}×2190\phantom{\rule{0ex}{0ex}}⇒1200=740x+2190y\phantom{\rule{0ex}{0ex}}⇒120=74x+219y....\left(2\right)$Step 5: Calculation of percentage of ${\mathrm{Mg}}^{+2}$ions.From equation (1) , $x+y=1$$⇒x=1–y\phantom{\rule{0ex}{0ex}}$Putting value of $x$ in equation 2: $⇒120=74\left(1–y\right)+219y\phantom{\rule{0ex}{0ex}}⇒120=74–74y+219y\phantom{\rule{0ex}{0ex}}⇒120=145y+74\phantom{\rule{0ex}{0ex}}⇒46=145y\phantom{\rule{0ex}{0ex}}⇒y=0.3172$Hence, the mass of ${\mathrm{Mg}}^{+2}$ ions is $0.3172\mathrm{g}$% of ${\mathrm{Mg}}^{+2}$ ions = $0.31721×100=31.72$Hence, $31.72%$ of ${\mathrm{Mg}}^{+2}$ ions is present.

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