Question

What is the mass of $2.18$ mol of Cu?

Answer 1

Hint : The mole is a measure or the base unit for the amount of substance present in the given sample. $1$ mole is a number that is equal to $6.022 \times {10^{23}}$ particles, also known as the Avogadro's constant. These particles can be any type of species, for example atoms, molecules, electrons, protons, neutrons etc.
The formula to calculate number of moles of a substance present in a given sample is as follows:
$Number{\text{ }}of{\text{ }}moles = \dfrac{{given{\text{ }}mass{\text{ }}of{\text{ }}substance}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}substance}}$

Complete Step By Step Answer:
We have been given, number of moles= $2.18$ mol of Cu
Molar mass of Cu= 63.5$$$$\dfrac{g}{{mol}}
Let given mass or mass of $2.18$mol of Cu be $x$
Now, putting the above values in the formula
$Number{\text{ }}of{\text{ }}moles = \dfrac{{given{\text{ }}mass{\text{ }}of{\text{ }}substance}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}substance}}$

$$2.18{\text{ }}mol = \dfrac{x}{{63.5g/mol}} \\\ x = 2.18{\text{ }}mol \times 63.5g/mol \\\ x = 138.43{\text{ }}g \\\$$

Hence, mass of $2.18$ mol of Cu is $138.43{\text{ }}g$
ADDITIONAL INFORMATION:
The mole unit is the base of the stoichiometry and it is providing the best option for expressing amounts of reactants and products consumed and formed during any chemical reaction.
The number of moles of a substance in a given pure sample can be represented by the following formula:
$n = \dfrac{N}{{{N_a}}}$
Where, $n$is the number of moles of the substance(or elementary entity)
$N$is the total number of elementary entities in the sample
${N_a}$ is the Avogadro’s Number which is numerically equal to $6.022 \times {10^{23}}$

Note :
Mass of one mole is equal to the molar mass of the substance, one should not get confused between the two. The number, $6.022 \times {10^{23}}$ (Avogadro's constant) is actually defined based on the number of atoms present in $12$g of $^{12}C$ isotope of Carbon. Where $12$gm is the mass of $1$ mole or $6.022 \times {10^{23}}$ atoms of $^{12}C$ isotope of Carbon.