Question

A solution containing 35 g of copper sulphate dissolved in 395 ml of water. Calculate the mass by mass percentage of the solution.

Answer 1

To describe a solution accurately, we are required to express its concentration i.e. how much of the solute has been dissolved in a specified amount of solvent. Words like dilute or concentrated are generally used to define the solutions that either consist of a little or a lot of amount of dissolved solute, respectively.

Complete step by step solution: Concentration is actually the measure of how much quantity of a given substance has been mixed with some other substance. There are different methods to express the concentration of solution by employing a percentage. The mass/mass percentage (i.e. % m/m) is stated as the ratio of the mass of a solute and the mass of a solution multiplied by 100:
$Mass{\text{ }}by{\text{ }}mass{\text{ }}percentage{\text{ }}of{\text{ }}a{\text{ }}solution = \dfrac{{Mass{\text{ }}of{\text{ }}solute}}{{Mass{\text{ }}of{\text{ }}solution}}X100$
We know that,
$mass{\text{ }}of{\text{ }}solution = mass{\text{ }}of{\text{ }}solute + mass{\text{ }}solvent$
You should note that each mass must possess the similar units in order to estimate the proper concentration.
In the given question, copper sulphate is the solute while water is the solvent.
Mass of solute ($CuS{O_4}$) = 35g (Given)
Volume of solvent (${H_2}O$) = 395 mL (Given)
We know that density of water = 1g/mL, that means:
Mass of solvent = 395 g
$\therefore mass{\text{ }}of{\text{ }}solution = 35 + 395 = 430g$
Now, substituting the values in %m/m of solution formula:
$Mass{\text{ }}by{\text{ }}mass{\text{ }}percentage{\text{ }}of{\text{ }}a{\text{ }}solution = \dfrac{{35g}}{{430g}}X100 = 8.14\%$

Hence, the mass by mass percentage of the solution is 8.14%.

Note: Another and the most common method to express concentration is the molarity. A solution's molarity tells you about the number of moles of solute that is present in 1 L of solution. You can say:
$Molarity(M) = \dfrac{{moles{\text{ }}of{\text{ solute}}}}{{1{\text{ }}litre{\text{ of }}solution{\text{ }}}}$