
How many megacoulombs of positive charge are in 1.00 mol of neutral molecular hydrogen gas $\left( {{{\text{H}}_2}} \right)$.
Answer
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Hint: One mole of any substance contains the avogadro number of entities. If one mole of hydrogen atom and another mole of hydrogen atom combine to form 1 mole of hydrogen molecule that means we have the Avogadro number of hydrogen molecules.
Complete answer:
Since it is given as a hydrogen molecule, each hydrogen molecule will be made of two individual hydrogen atoms. Each mole of hydrogen atom has Avogadro number of hydrogen atoms which literally means that each mole of hydrogen atom has Avogadro number of protons and Avogadro number of electrons. Since hydrogen atoms don't have any neutrons we are not concerned about them.
Value of Avogadro number is $6.022 \times {10^{23}}$
That means each mole of hydrogen molecule has $2 \times 6.022 \times {10^{23}}$ number of protons.
The magnitude of charge of an electron will be equal to the magnitude of charge of proton. Hence The magnitude of charge of proton will be $1.6 \times {10^{ - 19}}{\text{coulombs}}$
So the total amount of charge in coulombs in a neutral hydrogen molecule is $\eqalign{
& 2 \times 6.022 \times {10^{23}} \times 1.6 \times {10^{ - 19}}{\text{coulombs}} \cr
& \Rightarrow 19.27 \times {10^4}{\text{coulombs}} \cr
& \cr} $
We know that 1 megacoulomb (1MC) is equal to ${10^6}$ coulombs.
$\eqalign{
& 1{\text{MC = }}{10^6}{\text{coulombs}} \cr
& \Rightarrow {\text{1coulomb = }}\dfrac{1}{{{{10}^6}}}{\text{MC}} \cr
& \Rightarrow {\text{19}}.27 \times {10^4}{\text{coulombs = 19}}.27 \times {10^4} \times \dfrac{1}{{{{10}^6}}}{\text{MC}} \cr
& \Rightarrow {\text{19}}.27 \times {10^4}{\text{coulombs = 0}}{\text{.1927MC}} \cr} $
Hence 1 mole of neutral hydrogen molecule contains 0.1927 mega coulombs of positive charge.
Note:
The reason why they specifically mentioned neutral hydrogen atom is because to tell that there is no ionization occurred i.e no loss of electrons is happened. If oxidation or reduction had occurred then the molecule no longer remains neutral and then charge will be different.
Complete answer:
Since it is given as a hydrogen molecule, each hydrogen molecule will be made of two individual hydrogen atoms. Each mole of hydrogen atom has Avogadro number of hydrogen atoms which literally means that each mole of hydrogen atom has Avogadro number of protons and Avogadro number of electrons. Since hydrogen atoms don't have any neutrons we are not concerned about them.
Value of Avogadro number is $6.022 \times {10^{23}}$
That means each mole of hydrogen molecule has $2 \times 6.022 \times {10^{23}}$ number of protons.
The magnitude of charge of an electron will be equal to the magnitude of charge of proton. Hence The magnitude of charge of proton will be $1.6 \times {10^{ - 19}}{\text{coulombs}}$
So the total amount of charge in coulombs in a neutral hydrogen molecule is $\eqalign{
& 2 \times 6.022 \times {10^{23}} \times 1.6 \times {10^{ - 19}}{\text{coulombs}} \cr
& \Rightarrow 19.27 \times {10^4}{\text{coulombs}} \cr
& \cr} $
We know that 1 megacoulomb (1MC) is equal to ${10^6}$ coulombs.
$\eqalign{
& 1{\text{MC = }}{10^6}{\text{coulombs}} \cr
& \Rightarrow {\text{1coulomb = }}\dfrac{1}{{{{10}^6}}}{\text{MC}} \cr
& \Rightarrow {\text{19}}.27 \times {10^4}{\text{coulombs = 19}}.27 \times {10^4} \times \dfrac{1}{{{{10}^6}}}{\text{MC}} \cr
& \Rightarrow {\text{19}}.27 \times {10^4}{\text{coulombs = 0}}{\text{.1927MC}} \cr} $
Hence 1 mole of neutral hydrogen molecule contains 0.1927 mega coulombs of positive charge.
Note:
The reason why they specifically mentioned neutral hydrogen atom is because to tell that there is no ionization occurred i.e no loss of electrons is happened. If oxidation or reduction had occurred then the molecule no longer remains neutral and then charge will be different.
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