How many helium atoms are in a helium blimp containing 494kg of helium?
Answer
627.3k+ views
Hint: In the above question, we have to find the number of atoms present in 494kg of helium. So, basically, we have to count the number of moles of helium atoms present as compared with the Avogadro number. Avogadro number is the number of atoms present in $ {{12g}} $ of carbon-12 which is nearly equal to $ 6.022 \times {10^{23}} $ which comprises 1 mole of a substance.
Formula Used
$ {{n = }}\dfrac{{{m}}}{{{M}}} $
Where n is the number of moles
m is the given mass, and
M is the molar mass.
Complete step by step solution
We know that the atomic mass of He is 4.
Number of moles of helium = $ \dfrac{{{m}}}{{{M}}}{{ = }}\dfrac{{{{494000}}}}{{{4}}}{{ = 123500}} $
We know that : 1 mole of He = $ 6.022 \times {10^{23}} $ atoms
123500 moles of He = $ {{6}}{{.022 \times 1}}{{{0}}^{{{23}}}}{{ \times 123500 = 7}}{{.43 \times 1}}{{{0}}^{{{28}}}} $
Therefore, $ {{7}}{{.43 \times 1}}{{{0}}^{{{28}}}} $ helium atoms are present in 494kg of helium.
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:
$ {{1 mole}} $ = $ 6.022 \times {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
$ {{n = }}\dfrac{{{m}}}{{{M}}} $
Where n is the number of moles
m is the given mass, and
M is the molar mass.
Complete step by step solution
We know that the atomic mass of He is 4.
Number of moles of helium = $ \dfrac{{{m}}}{{{M}}}{{ = }}\dfrac{{{{494000}}}}{{{4}}}{{ = 123500}} $
We know that : 1 mole of He = $ 6.022 \times {10^{23}} $ atoms
123500 moles of He = $ {{6}}{{.022 \times 1}}{{{0}}^{{{23}}}}{{ \times 123500 = 7}}{{.43 \times 1}}{{{0}}^{{{28}}}} $
Therefore, $ {{7}}{{.43 \times 1}}{{{0}}^{{{28}}}} $ helium atoms are present in 494kg of helium.
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:
$ {{1 mole}} $ = $ 6.022 \times {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.
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