Question

How many grams of NaCl are in 225 ml of a 0.75M NaCl solution? (The molecular weight of NaCl is 58.44 g.)

Answer 1

Hint: This question can be solved using the mole concept and the concept of molarity. The molarity of a solution is defined as the concentration of a solute in a solution is in terms of the amount of solute per unit volume, or the molar concentration.

Complete answer:
Molarity can be given by
\[M=\dfrac{{{w}_{s}}}{{{M}_{s}}}\times \dfrac{1000}{V}\]
Where ${{w}_{s}}$= mass of the solute dissolved (in grams),
${{M}_{s}}$= molar mass of the solute dissolved (in grams),
And V= total volume of the solution (in mL).
The SI unit for molarity is mol/${{m}^{3}}$ but it is usually expressed in mol/$d{{m}^{3}}$ which is equivalent to mol/L. It can also be expressed by the capital letter M.
Now, the molecular mass of one mole of a compound is equal to the sum of atomic masses of all the elements present in the molecule. We know that the molecular weight of NaCl is 58.44 g.
It is given to us that the molarity of the sodium chloride solution M=0.75 mol/L and the volume of the sodium chloride solution is V=225 mL.
Upon substituting these values in the molarity equation, we get
\[\begin{align}
  & M=\dfrac{{{w}_{NaCl}}}{{{M}_{NaCl}}}\times \dfrac{1000}{V} \\
 & {{w}_{NaCl}}=\dfrac{M\times {{M}_{NaCl}}\times V}{1000} \\
 & {{w}_{NaCl}}=\dfrac{0.75\times 58.44\times 225}{1000} \\
 & {{w}_{NaCl}}\cong 9.86g \\
\end{align}\]
So, there are approximately 9.86 grams of NaCl in 225 ml of a 0.75M NaCl solution

Note:
It should be noted that if one mole of a substance is present, it has exactly the Avogadro number (${{N}_{A}}=6.022\times {{10}^{23}}$) of particles and the mass of one mole of a compound is equivalent to the sum of the mass of all the particles contained in one mole of a substance. This can also be used to define the molar mass of a substance.