
The number of electrons in 2 gram of nitrate ions $\left( \text{N}{{\text{O}}_{\text{3}}}^{\text{-}} \right)$is:
A.$\text{64 }{{\text{N}}_{\text{A}}}$
B.${{\text{N}}_{\text{A}}}$
C.32
D.$\text{32 }{{\text{N}}_{\text{A}}}$
Answer
563.4k+ views
Hint:${{\text{N}}_{\text{A}}}$or the Avogadro’s Number is defined as the number of particles present in one mole of a substance and is equal to $6.023\times {{10}^{23}}$particles. Here the word “particles” refer to ions, atoms, molecules, ions, and subatomic particles.
Complete step by step answer:
To find out the number of in 2 gram of nitrate ions $\left( \text{N}{{\text{O}}_{\text{3}}}^{\text{-}} \right)$, we need to find out the gram ionic weight of nitrate.
1 mole of nitrate ions = $\left[ 14+\left( 16\times 3 \right) \right]=62$ grams = four atoms (one atom of nitrogen and three atoms of oxygen)
In one atom of oxygen, there are eight electrons and in one atoms of nitrogen there are seven electrons, and one electron from the charge of the nitrate, hence the total number of electrons in the nitrate anion = $\left[ 8+\left( 7\times 3 \right)+1 \right]=30$ electrons.
So one unit of nitrate contains 30 electrons, therefore,
$6.023\times {{10}^{23}}$ Nitrate particles must contain\[\text{6}\text{.023 }\times\text{ 1}{{\text{0}}^{\text{23}}}\text{ }\times\text{ 30 = 30 }{{\text{N}}_{\text{A}}}\] electrons.
Now, 62 grams of nitrate = ${{\text{N}}_{\text{A}}}$Nitrate particles = \[\text{30 }{{\text{N}}_{\text{A}}}\]electrons.
Therefore, 2 grams of nitrate = $\dfrac{\text{2}}{\text{62}}\text{ }\!\!\times\!\!\text{ 30 }{{\text{N}}_{\text{A}}}=\dfrac{30}{31}{{\text{N}}_{\text{A}}}\approx {{\text{N}}_{\text{A}}}$ electrons.
The correct answer is therefore option B, 2 grams of nitrate ions contains Avogadro’s number of electrons.
Note:
The ionic weight of an ionic species is defined as the weight in grams of an ion that is equal to the sum of the atomic weights of the atoms making up the ionic species. This is also known as one gram ion. One gram ion is also equal to the amount of ions that contain the Avogadro number of ions, i.e., $6.023\times {{10}^{23}}$ ions of a substance or one mole of ions.
Complete step by step answer:
To find out the number of in 2 gram of nitrate ions $\left( \text{N}{{\text{O}}_{\text{3}}}^{\text{-}} \right)$, we need to find out the gram ionic weight of nitrate.
1 mole of nitrate ions = $\left[ 14+\left( 16\times 3 \right) \right]=62$ grams = four atoms (one atom of nitrogen and three atoms of oxygen)
In one atom of oxygen, there are eight electrons and in one atoms of nitrogen there are seven electrons, and one electron from the charge of the nitrate, hence the total number of electrons in the nitrate anion = $\left[ 8+\left( 7\times 3 \right)+1 \right]=30$ electrons.
So one unit of nitrate contains 30 electrons, therefore,
$6.023\times {{10}^{23}}$ Nitrate particles must contain\[\text{6}\text{.023 }\times\text{ 1}{{\text{0}}^{\text{23}}}\text{ }\times\text{ 30 = 30 }{{\text{N}}_{\text{A}}}\] electrons.
Now, 62 grams of nitrate = ${{\text{N}}_{\text{A}}}$Nitrate particles = \[\text{30 }{{\text{N}}_{\text{A}}}\]electrons.
Therefore, 2 grams of nitrate = $\dfrac{\text{2}}{\text{62}}\text{ }\!\!\times\!\!\text{ 30 }{{\text{N}}_{\text{A}}}=\dfrac{30}{31}{{\text{N}}_{\text{A}}}\approx {{\text{N}}_{\text{A}}}$ electrons.
The correct answer is therefore option B, 2 grams of nitrate ions contains Avogadro’s number of electrons.
Note:
The ionic weight of an ionic species is defined as the weight in grams of an ion that is equal to the sum of the atomic weights of the atoms making up the ionic species. This is also known as one gram ion. One gram ion is also equal to the amount of ions that contain the Avogadro number of ions, i.e., $6.023\times {{10}^{23}}$ ions of a substance or one mole of ions.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a labelled diagram of the human heart and label class 11 biology CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

