Question

When 10 mL of ethyl alcohol (density = 0.7893 g/mL) is mixed with 20 mL of water (density 0.9971 g/mL) at 25 C, the final solution has a density of 0.9571 g/mL. The percentage change in total volume on mixing is:
A. 3.1 %
B. 2.4 %
C. 1 %
D. None of these.

Answer 1

Hint: Mass of a chemical is nothing but the product of density and volume of the chemical.
- We can calculate the percentage of change in total volume on mixing by using the following formula.
\[\begin{align}
& \text{The percentage of change in total volume on mixing is =} \\
& \dfrac{\text{total volume of ethanol and water before mixing-volume of solution after mixing}}{\text{total volume of ethanol and water before mixing}}\times 100 \\
\end{align}\]

Complete Solution :
- We have to calculate the percentage change in total volume on mixing of two solutions.
- The given density of the ethyl alcohol = 0.7893 g/mL
- The given volume of the ethyl alcohol = 10 mL
- Then the Mass of the ethyl alcohol = ( Density) (Volume) = (0.783) (10) = 7.803 g.
- The given density of the water = 0.9971 g/mL
- The given volume of the ethyl alcohol = 20 mL
- Then the Mass of the water = ( Density) (Volume) = (0.9971) (20) = 27.835 g.
- Volume of solution after mixing water and ethyl alcohol is as follows.
$\dfrac{Mass}{Density} = \dfrac{27.835}{0.9571} = 29.083mL$

- Then the total volume of the solution = volume of ethanol + Volume of water = 10 +20 = 30 mL.
- Now substitute all the known values in the below equation to get the answer.
\[\begin{align}

& \text{The percentage of change in total volume on mixing is =} \\
& \dfrac{\text{total volume of ethanol and water before mixing - volume of solution after mixing}}{\text{total volume of ethanol and water before mixing}}\times 100 \\
 & =\dfrac{30 - 29.083}{30}\times 100 \\
 & = 3.1 \\
\end{align}\]
- The percentage change in total volume on mixing of two solutions is 3.1 %.
So, the correct answer is “Option A”.

Note: If we are going to add a volatile solvent to a non-volatile solvent then the volume of the solution will be lesser than the volume of the solvents added individually due to the loss of volatile solvent in the form of evaporation.