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?

Answer 1

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.

$\Rightarrow x+y=1....\left(1\right)$ as total mass is 1g

$\Rightarrow MolesofM{g}^{+}ions=\frac{x}{24}$

$\Rightarrow MolesofM{g}^{+2}ions=\frac{y}{24}$

Step 4: Calculation of energy absorbed.

The energy absorbed for generating $\frac{x}{24}$moles of ${\mathrm{Mg}}^{+}$ ions =$\frac{x}{24}\times 740kJmo{l}^{-1}$

The energy absorbed for generating $\frac{y}{24}$moles of ${\mathrm{Mg}}^{+2}$ ions= $\frac{y}{24}\times 2190kJmo{l}^{-1}$

Total energy absorbed =$\left(\frac{x}{24}\times 740+\frac{y}{24}\times 2190\right)kJmo{l}^{-1}$

$$\begin{gathered} \Rightarrow 50=\frac{x}{24}\times 740+\frac{y}{24}\times 2190 \\ \Rightarrow 1200=740x+2190y \\ \Rightarrow 120=74x+219y....\left(2\right) \end{gathered}$$

Step 5: Calculation of percentage of ${\mathrm{Mg}}^{+2}$ions.

From equation (1) ,

$x+y=1$

$\Rightarrow x=1–y$

Putting value of $x$ in equation 2:

$$\begin{gathered} \Rightarrow 120=74(1–y)+219y \\ \Rightarrow 120=74–74y+219y \\ \Rightarrow 120=145y+74 \\ \Rightarrow 46=145y \\ \Rightarrow y=0.3172 \end{gathered}$$

Hence, the mass of ${\mathrm{Mg}}^{+2}$ ions is $0.3172\mathrm{g}$

% of ${\mathrm{Mg}}^{+2}$ ions = $0.31721\times 100=31.72$

Hence, $31.72\%$ of ${\mathrm{Mg}}^{+2}$ ions is present.