Question

Calculate the energy required to convert all atoms in $4.8\, g$ of $Mg$ to $M g^{2+}$ in the vapor state. $I E_{1}$ and $I E_{2}$ of $M g$ are $740\, kJ / mol$ and $1450\, kJ / mol$ respectively.

• A. +740 kJ / mol
• B. -740 kJ / mol
• C. -1450 kJ / mol
• D. +438 kJ / mol

Answer 1

Answer: (D)

For $1$ mole of $\left(M g \rightarrow M g^{2+}\right)$
$I E=I E_{1}+I E_{2}=(740+1450)=2190\, kJ / mol$
Number of mole in $4.8\, g$ of $M g=4.8 / 24=0.2\, mol$
For $1$ mole energy required $=2190\, kJ / mol$
For $0.2$ mole energy required $=2190 \times 0.2=438\, kJ$
Thus, for $4.8\, g$ of $M g$ to $M g^{2+}$ conversion, energy required is $438\, kJ$.