Find the charge in coulomb on 1 gram ion of ${{N}^{3-}}$.
Answer
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Hint: As we are supposed to find the charge on one gram of nitride ion so, we will consider its oxidation number along with the charge on electron and Avogadro’s number which are generally for every type of ion.
Formula used: $\text{Oxidation number}\times \text{charge on electron}\times \text{Avogadro }\!\!'\!\!\text{ s number}$
Complete answer:
Ion: An ion has a tendency of bearing a charge of more than one. It can be positive or negative depending upon the sign on an atom. We call an atom carrying a positive charge as cation whereas an atom carrying a negative charge is anion.
Nitride: The word nitride is similar to the word nitrogen but there is a difference between the two. The difference is about their oxidation states. Nitride has an oxidation state of -3 whereas nitrogen has no oxidation state rather it has atomic number and atomic weight whose values are 7 and 140067 respectively.
As we know that the charge on an electron is $1.6\times {{10}^{-19}}C$. Thus, the charge on one ion will be $3\times 1.6\times {{10}^{-19}}C$.
Also, as we know that the Avogadro’s number represents the number of moles in one mole of a particular substance so, 1 gram of the negative ion or anion will be $6.022\times {{10}^{23}}C$ ions.
By using all the above information we will now find the charge on 1 gram of this ion which is nitride. Therefore, we get
$\begin{align}
& \text{Oxidation number}\times \text{charge on electron}\times \text{Avagadro }\!\!'\!\!\text{ s number} \\
& \Rightarrow 3\times 1.6\times {{10}^{-19}}\times 6.022\times {{10}^{23}}=2.89\times {{10}^{5}}C \\
\end{align}$.
Note:
There is a use of Avogadro’s number whenever 1 gram of mole is used. Here, the charge of an ion namely anion which is nitride is to be found here. For any anion if we need to get the charge on 1 gram of any substance so, we will include Avogadro’s number and the basic charge on an ion. These numbers are very important for numerical. As these numbers are big, we should practice them by practicing on paper. Be cautious while solving the numerical. If the numerical is not solved with focus then, it will lead to wrong answers.
Formula used: $\text{Oxidation number}\times \text{charge on electron}\times \text{Avogadro }\!\!'\!\!\text{ s number}$
Complete answer:
Ion: An ion has a tendency of bearing a charge of more than one. It can be positive or negative depending upon the sign on an atom. We call an atom carrying a positive charge as cation whereas an atom carrying a negative charge is anion.
Nitride: The word nitride is similar to the word nitrogen but there is a difference between the two. The difference is about their oxidation states. Nitride has an oxidation state of -3 whereas nitrogen has no oxidation state rather it has atomic number and atomic weight whose values are 7 and 140067 respectively.
As we know that the charge on an electron is $1.6\times {{10}^{-19}}C$. Thus, the charge on one ion will be $3\times 1.6\times {{10}^{-19}}C$.
Also, as we know that the Avogadro’s number represents the number of moles in one mole of a particular substance so, 1 gram of the negative ion or anion will be $6.022\times {{10}^{23}}C$ ions.
By using all the above information we will now find the charge on 1 gram of this ion which is nitride. Therefore, we get
$\begin{align}
& \text{Oxidation number}\times \text{charge on electron}\times \text{Avagadro }\!\!'\!\!\text{ s number} \\
& \Rightarrow 3\times 1.6\times {{10}^{-19}}\times 6.022\times {{10}^{23}}=2.89\times {{10}^{5}}C \\
\end{align}$.
Note:
There is a use of Avogadro’s number whenever 1 gram of mole is used. Here, the charge of an ion namely anion which is nitride is to be found here. For any anion if we need to get the charge on 1 gram of any substance so, we will include Avogadro’s number and the basic charge on an ion. These numbers are very important for numerical. As these numbers are big, we should practice them by practicing on paper. Be cautious while solving the numerical. If the numerical is not solved with focus then, it will lead to wrong answers.
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