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Ionization energy (IE) of Na is 513 kJ/mol. The value in eV/molecule is $x\text{ x}{{10}^{-1}}$. Then the value of x is ___________.

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Hint: Ionization energy (IE) is a measure of the amount of energy required to take out an electron from the shell of an atom. Energy is supplied to surpass the existing forces of attraction in the element. It is the highest for noble gases.
Complete answer:
Let us understand the concept of Ionization Potential before the calculations.
“The first ionization energy is the energy required to remove one mole of the most loosely held electrons from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of +1.” It can be portrayed by the following reaction –
$M(g)\to {{M}^{+}}(g)+{{e}^{-}}$
The amount of energy which is required to carry out the above reaction is known as ionization potential.
Ionization energy can be expressed in any of the following units –
eV/ molecule
kJ/ mole
kcal/ mole
Now, according to the question, Ionization energy (IE) of Na is 513 kJ/mol.
1 kJ/mole = $\dfrac{1000}{{{N}_{A}}}$ J/ molecule
Where, ${{\text{N}}_{\text{A}}}$ = Avogadro’s number
513 kJ/mole = 513 x $\dfrac{1000}{{{N}_{A}}}$ J/ molecule = 513 x $\dfrac{\text{1000}}{\text{6}\text{.02 x 1}{{\text{0}}^{\text{23}}}}\text{ = 8}\text{.522 x 1}{{\text{0}}^{\text{-19}}}$ J/ molecule

1 Joule = $\dfrac{\text{1}}{\text{1}\text{.6 x 1}{{\text{0}}^{\text{19}}}}$eV
So, we get $\dfrac{\text{8}\text{.522x1}{{\text{0}}^{\text{-19}}}}{\text{1}\text{.6x1}{{\text{0}}^{\text{19}}}}$eV = $\text{53}\text{.26 x}{{10}^{-1}}$ eV.

Therefore, the answer is – The value of x is 53.26.

Additional Information: Noble gases have the highest ionization energies because they are very stable.

Note: Alternately, you can also do it by remembering a simple conversion –
1 eV = 96.49 kJ/mol.
Given, Ionization energy (IE) of Na is 513 kJ/mol.
From the above conversion, 1 kJ/ mol = $\dfrac{1}{96.49}$ eV
Therefore, 513 kJ/ mol = $\dfrac{513}{96.49}$ eV = 5.32 eV
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