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Find the energy in $ kJ $ per mole of electronic charge accelerated by a potential of $ 2 $ volts.

Answer
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Hint: To solve this question using the formula for the conversion of joules to electron volt. Then find the value of energy in $ kJ $ per mole by calculating the total charge present in one mole of electrons using the value of Avogadro number to find the number of particles present in one mole of matter. Keep in mind to convert $ J $ to $ kJ $ by dividing the value by $ {10^3} $ .

Complete step by step answer:
We need to find the energy in the unit $ kJ $ per mole.
Using Avogadro's number we know that $ 1 $ mole of electrons contains $ 6.02 \times {10^{23}} $ number of electrons.
We know that the charge present in $ 1 $ electrons is equal to $ 1.6 \times {10^{ - 19}}C $ .
Therefore the total charge in coulombs $ (C) $ of $ 1 $ mole of electron can be given as
 $ = (1.6 \times {10^{ - 19}} \times 6.02 \times {10^{23}})C $
It is given that the electronic charge is accelerated by a potential of $ 2 $ volts $ (V) $ .
We know that energy in joules can be calculated by multiplying the potential difference present in volts $ (V) $ with the charge of an electron ( $ e $ ).
Hence, $ 1J = 1eV $
 $ \therefore 1kJ = {10^3}J = {10^3}eV $
Therefore, upon substituting the values we can say that,
Energy in $ kJ/mol = \dfrac{{(1.6 \times {{10}^{ - 19}} \times 6.02 \times {{10}^{23}})C \times 2V}}{{{{10}^3}}} $
 $ \Rightarrow $ Energy in $ kJ/mol = \dfrac{{192640}}{{{{10}^3}}} $
 $ \Rightarrow $ Energy in $ kJ/mol = 192.64 \approx 193 $
Therefore the nearest integer of the value of energy in $ kJ $ per mole of electronic charge accelerated by a potential of $ 2 $ volts is $ 193kJ $ per mole.

Note:
Even though joule $ (J) $ is the SI unit of energy, there are other units of energy as well such as electron volt $ (eV) $ , kiloWatt-hour $ (kWh) $ , calorie $ (cal) $ , erg $ (erg) $ , and so on. All of these units of energy as inter convertible to one another. We can use any of these units according to the requirement of the given question.