Question
Question: How many atoms are present in \(4.00\) mol of aluminum ?...
How many atoms are present in 4.00 mol of aluminum ?
Solution
With the help of Avogadro's no. we can determine mol to atoms conversion easily using unit conversion techniques, we can use the mole labeled to convert back and forth between the number of particles and moles. Most Important values here to determine atoms from mole is Avogadro's number, 6.022×1023
Complete answer:
For mol to mass:
4 mol of aluminum atoms
Mass of 1 mol of aluminum atoms =27g
Mass of 4 mol of aluminum atoms = (27 × 4) = 108g
For mol to atom:
To determine the amount of atoms (or molecules) present given the moles of an element, you multiply the amount of moles by Avogadro's number, 6.022×1023
4 mol of Al ×(1mol6.022×1023) = 2.41 ×102 atoms/mol of Al
Therefore, 2.41×102 atoms/mol of Aluminum are present in 4 mol of Aluminum.
Additional Information: Why is Avogadro's number referred to as a mole?
In calculations, sometimes we have to have conversions which are necessary in order to calculate and determine the desired answer. A conversion can be performed by knowing the relationship between two variables. Avogadro's number is used for conversion between molecules or atoms to moles. It shows the relationship between the number of particles such as atoms and molecules to the mole of a compound or substance.
The relationship of Avogadro's number to moles is: 6.022×1023
It is referred to as a mole since it shows the number of particles such as atoms and molecules that are present per mole of a given compound or substance. It provides a distinct numerical value of the particles associated with one mole of the compound or substance.
Note: We have to consider Avogadro's number is a measured quantity with three significant figures; the result of the calculation is rounded to three significant figures. The first conversion factor converts from moles of particles to the number of particles. The second conversion factor reflects the number of atoms contained within each molecule.