How many silver atoms are there in $ {\text{4}}{\text{.01}} $ g of silver?
Answer
566.4k+ views
Hint: In the above question, we are asked to find out the number of silver atoms present in the given amount of silver. For this, we have to first find out the number of moles of silver atoms present. We know that 1 mole of silver contains $ {{\text{N}}_{\text{A}}} $ atoms and hence, after comparing we will get the number of silver atoms present.
Formula used:
$ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}} $ , where
n= number of moles
m= given mass
M= molar mass .
Complete step by step solution:
Let us first find out the number of moles of silver present.
Molar mass of silver is 108.
So, $ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}{\text{ = }}\dfrac{{{\text{4}}{\text{.01}}}}{{{\text{108}}}}{\text{ = 0}}{\text{.037}} $
$ 1 $ mole of the Ag $ = $ $ {{\text{N}}_{\text{A}}} $ atoms
Now, to find out the number of atoms present in $ {\text{0}}{\text{.037}} $ mole of the silver, we will multiply $ 0.037 $ to both sides of the equations.
So, $ {\text{0}}{\text{.037}} $ of the silver $ = $ $ {\text{0}}{\text{.25 }}{{\text{N}}_{\text{A}}}{\text{ = 0}}{\text{.037 6}}{\text{.022 1}}{{\text{0}}^{{\text{23}}}}{\text{ = 2}}{\text{.22 1}}{{\text{0}}^{{\text{22}}}} $ atoms
$ \therefore $ There are $ {\text{2}}{\text{.22 1}}{{\text{0}}^{{\text{22}}}} $ silver atoms in $ {\text{4}}{\text{.01}} $ g of silver.
Additional Information
Avogadro’s number is used in chemistry to deal with large numbers. It is the basis of the mole unit of measurement. It is a simple way of conversion between mole, mass and number of molecules. Suppose, to get the number of particles present in a substance, first, we have to find the number of moles present in it and then by using Avogadro’s number we can find the number of particles present inside it.
Note:
Avogadro’s number is kind of generalized number, hence, we can write:
$ {\text{1 mole}} $ = $ 6.022 {10^{23}} $ atoms, or molecules, or protons, or electrons etc.
In short, it is the number of particles in a mole.
Hence, in order to solve these types of questions, we have to first decide which quantity (electron, proton, atom) is equivalent to Avogadro’s number.
Formula used:
$ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}} $ , where
n= number of moles
m= given mass
M= molar mass .
Complete step by step solution:
Let us first find out the number of moles of silver present.
Molar mass of silver is 108.
So, $ {\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}{\text{ = }}\dfrac{{{\text{4}}{\text{.01}}}}{{{\text{108}}}}{\text{ = 0}}{\text{.037}} $
$ 1 $ mole of the Ag $ = $ $ {{\text{N}}_{\text{A}}} $ atoms
Now, to find out the number of atoms present in $ {\text{0}}{\text{.037}} $ mole of the silver, we will multiply $ 0.037 $ to both sides of the equations.
So, $ {\text{0}}{\text{.037}} $ of the silver $ = $ $ {\text{0}}{\text{.25 }}{{\text{N}}_{\text{A}}}{\text{ = 0}}{\text{.037 6}}{\text{.022 1}}{{\text{0}}^{{\text{23}}}}{\text{ = 2}}{\text{.22 1}}{{\text{0}}^{{\text{22}}}} $ atoms
$ \therefore $ There are $ {\text{2}}{\text{.22 1}}{{\text{0}}^{{\text{22}}}} $ silver atoms in $ {\text{4}}{\text{.01}} $ g of silver.
Additional Information
Avogadro’s number is used in chemistry to deal with large numbers. It is the basis of the mole unit of measurement. It is a simple way of conversion between mole, mass and number of molecules. Suppose, to get the number of particles present in a substance, first, we have to find the number of moles present in it and then by using Avogadro’s number we can find the number of particles present inside it.
Note:
Avogadro’s number is kind of generalized number, hence, we can write:
$ {\text{1 mole}} $ = $ 6.022 {10^{23}} $ atoms, or molecules, or protons, or electrons etc.
In short, it is the number of particles in a mole.
Hence, in order to solve these types of questions, we have to first decide which quantity (electron, proton, atom) is equivalent to Avogadro’s number.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

