Question

How many moles of \[NaCl\;\] are present in \[500{\text{ }}mL\] of a \[1.55\;M\] \[NaCl\] solution?

Answer 1

Hint: The mole concept is very significant and useful in chemistry. It is actually the base of stoichiometry and it provides the best option to express the amounts of reactants as well as products that are consumed and formed during a chemical reaction.

Complete step by step answer:
To calculate the number of moles, we generally use the following formula:
$Number{\text{ }}of{\text{ }}moles = \dfrac{{Mass(g)}}{{Molar{\text{ }}mass(gmo{l^{ - 1}})}}$
In the given question we are not provided with the mass of \[NaCl\] but its molarity is given. So, it means we cannot use the above mentioned formula to calculate the number of moles.
The information that is being provided to us in the question is stated below:
Volume$ = 500{\text{ }}mL = 0.5{\text{ }}L$(We have converted the units to liters by dividing it with 1000)
Molarity = \[1.55\;M\]
We know that solution's molarity tells you about the number of moles of solute that is present in 1 L of solution. You can use the following formula for molarity calculation:
$Molarity(M) = \dfrac{{moles{\text{ }}of{\text{ solute}}}}{{1{\text{ }}litre{\text{ of }}solution{\text{ }}}}$
Now, substituting the given values in the aforementioned formula of molarity, we can calculate the number of moles of \[NaCl\;\] by rearranging the equation as shown below:
$
  1.55 = \dfrac{{moles{\text{ }}of{\text{ }}NaCl}}{{0.5{\text{ }}}} \\
  moles{\text{ }}of{\text{ }}NaCl = 0.775 \\
$

Hence, 0.775 moles of \[NaCl\;\] are present in \[500{\text{ }}mL\] of a \[1.55\;M\] \[NaCl\] solution.

Note: Don’t get confused between molarity and molality. Molarity is the number of moles of solute per volume of solution (litres) while Molality is the number of moles of solute per weight of solvent (kilogram). These two have a relationship as below:
Molality = (density of the solution − molarity) $\times$ molecular weight of solute molarity