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Question: Find the charge on 1g-ion of \({{N}^{3-}}\) in coulombs....

Find the charge on 1g-ion of N3{{N}^{3-}} in coulombs.

Explanation

Solution

Hint : Take into consideration the definition of what a gram-ion is and then associate it with the charge of 1 mole of electrons and carry out calculations accordingly.

Complete step by step solution:
A gram-ion means the number of ions that are present in 1 mole of the substance. NN has only 1 gram-ion since it forms only one ion i.e., N3{{N}^{3-}} at a time.
Therefore, 1g-ion of N3{{N}^{3-}} means 1 mole of N3{{N}^{3-}} is present. There are 3 electrons in this ion, hence, 3 moles of electrons are present in 1g-ion or 1 mole of N3{{N}^{3-}}.
The charge present on 1 mole of electrons has already been defined by Faraday, it is known as the Faraday’s constant and it is denoted by FF. The accepted value of this constant is 96485.33212...Cmol196485.33212...Cmo{{l}^{-1}} but we will round it off to 96500 for ease in calculation.
Charge present on 1 mole of electrons = 9650096500 C
Charge present on 3 moles of electrons = 96500×396500\times 3 C
Charge on 1g-ion of N3{{N}^{3-}} = 2.895×1052.895\times {{10}^{5}} C
Therefore, the answer to this question is 2.895×1052.895\times {{10}^{5}}C

Additional information:
Another method to solve this problem is to consider the charge on 1 ion of N3{{N}^{3-}}.
Charge on 1 ion of N3{{N}^{3-}}= 3×1.6×10193\times 1.6\times {{10}^{-19}} C
Where 1.6×10191.6\times {{10}^{-19}} C is the charge present on 1 electron
Charge on 1 ion of N3{{N}^{3-}}= 4.8×10194.8\times {{10}^{-19}} C
Now, we will multiply the charge present on 1 ion of N3{{N}^{3-}} with the number of ions that are present in 1g-ion of N3{{N}^{3-}}. We already know that 1g-ion of N3{{N}^{3-}} means 1mol of N3{{N}^{3-}} is present. The number of atoms/ions that are present in 1mol of a substance is 6.022×10236.022\times {{10}^{23}} which is Avogadro's number. Hence, we will multiply the charge present on 1 ion of N3{{N}^{3-}} by Avogadro's number.
Charge on 1 mole of N3{{N}^{3-}}ions = charge on 1 ion of N3×{{N}^{3-}}\times number of ions
Charge on 1 mole of N3{{N}^{3-}}ions = (4.8×1019)×(6.022×1023)(4.8\times {{10}^{-19}})\times (6.022\times {{10}^{23}}) C
Charge on 1 mole of N3{{N}^{3-}}ions = 2.89×1052.89\times {{10}^{5}} C

Note : Please do not get confused between 1 gram of X ions and 1g-ion of X. Here, X refers to N3{{N}^{3-}}. Remember that 1g-ion means the number of ions present in 1 mole of a substance. For example, ZnSO4ZnS{{O}_{4}}, contains 2g-ions. One of Zn2+Z{{n}^{2+}} and one of SO42S{{O}_{4}}^{2-}. The charge can also be written in terms of Faraday in any problem, in this problem the charge will be 3 Faraday. The charge on 1 mole of electrons is 1 Faraday.