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Question

(D) The total ionization energy required to convert $1 \ mol$ of $Mg$ atoms to $Mg^{2+}$ ions is the sum of the first and second ionization energies:

Vedclass · Accepted Answer

$+438 \ kJ$

Vedclass · Answer

(D) The total ionization energy required to convert $1 \ mol$ of $Mg$ atoms to $Mg^{2+}$ ions is the sum of the first and second ionization energies: $IE_{total} = IE_1 + IE_2 = 740 \ kJ/mol + 1450 \ kJ/mol = 2190 \ kJ/mol$. The number of moles in $4.8 \ g$ of $Mg$ (atomic mass $= 24 \ g/mol$) is calculated as: $n = \frac{4.8 \ g}{24 \ g/mol} = 0.2 \ mol$. The total energy required for $0.2 \ mol$ of $Mg$ is: $E = n 	imes IE_{total} = 0.2 \ mol 	imes 2190 \ kJ/mol = 438 \ kJ$. Thus,the energy required is $438 \ kJ$. Hence,the correct option is $(D)$.

Calculate the energy required to convert all atoms in $4.8 \ g$ of $Mg$ to $Mg^{2+}$ in the vapour state. $IE_1$ and $IE_2$ of $Mg$ are $740 \ kJ / mol$ and $1450 \ kJ / mol$ respectively.

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