Question
Question: For \(26.8\) moles of argon contain how many atoms?...
For 26.8 moles of argon contain how many atoms?
Solution
We need to remember that the argon is one of the elements in the periodic table. It is one of the inert gas in the world. Its atomic number is 18 . The symbol of argon is Ar . The mole is one of the chemical units used to measure the molecules. It is mainly used to express the quantity of individual atoms. One mole means the amount of substance, which contains elementary particles. Here we considered atoms in 12g of carbon- 12 isotope. These elementary particles include molecules, atoms, ions, electrons or other particles. By using this mole concept, we find the number of atoms of the element and the number of molecules in the compound.
Formula used:
The number of atoms in the atom is equal to the product of the number of moles of the atom and Avogadro’s number.
The Avogadro’s number is 6.022×1023. It is represented by NA.
Apply, Avogadro’s value we get the formula
Number of atoms= Number of moles ×6.022 ×1023
Complete step by step answer:
Given,
The number of moles in argon = 26.8 moles.
Calculate the number of atoms of argon present in given moles,
Number of atoms= Number of moles ×6.022 ×1023
1 mole of Ar = 1×6.022×1023
=6.022×1023
One mole of argon having 6.022×1023 atoms of argon.
26.8moles of argon is calculated,
26. mole of Ar = 26.8×6.022×1023
=161.3896×1023
=1.613896×1025
Hence, 26.8moles of argon is having 1.613896×1025 atoms of argon.
Note:
This mole unit plays an important role in chemical reaction. By using this mole only we concluded the reaction, how moving. Molar mass is the concept, the mass of an atom present in one mole of substance. Avogadro’s number doesn’t have any units. Avogadro’s number is named after Amedeo Avogadro. He is Italian scientist. He proposed the equal volume of all the gases molecules under the same condition of temperature and pressure contain equal numbers of molecules in the same substance. This is the reason for keeping his name for this number.