Question

$10mL$ of ${H_2}S{O_4}$ is dissolved in $90mL$ of water. Calculate mass by volume percentage of solution.

Answer 1

Hint : First calculate the mass of sulphuric acid dissolved in water. Use the relation of density with mass and volume. The density of ${H_2}S{O_4}$ is $1.84g/mL$ . To get the mass by volume percentage of the solution, we will divide the mass of solution by the volume of the solution and finally multiply it by $100$ . This multiplication is done to get the percentage.

Complete Step By Step Answer:
The mass by volume percent of a solution is defined as the ratio of the mass of solute that is present in a solution, relative to the volume of the solution, as a whole. Because this type of concentration is expressed as a percentage, the indicated proportion must be multiplied by $100$ as given below:
Mass by volume percentage of solution $= \dfrac{{mass{\text{ }}of{\text{ }}solution}}{{volume{\text{ }}of{\text{ }}solution}} \times 100$
First, let’s calculate the mass of ${H_2}S{O_4}$ in $10mL$ solution. We will use the relation of density with mass and volume.
We know that,
Mass $=$ density $\times$ volume
$= 10mL \times 1.84g/mL$
$= 18.4g$
The total volume of the entire solution is $(10 + 90)mL = 100mL$
Now, let’s input all these values into the formula to get our solution.
Mass by volume percentage of solution $= \dfrac{{mass{\text{ }}of{\text{ }}solution}}{{volume{\text{ }}of{\text{ }}solution}} \times 100$
$= \dfrac{{18.4g}}{{100mL}} \times 100$
$= 18.4\%$
Therefore, the mass by volume percentage of the solution is $18.4\%$ .

Note :
In the question, if in place of volume, mass is given along with density, then to get the volume of solution, you need to divide the density by mass.
The mass by volume percentage of a solution is also known as mass concentration of a solution as it tells us about the mass of solute present in the solution. If the mass of solute is present in a relatively large amount then it will be a concentrated solution. On the other hand, if the mass of solute is present in a low amount, then it will be a dilute solution.