Question

The specific gravity of a salt solution is 1.025. If $V$ml of water is added to 1.0 L of this solution to make its density 1.02g $m{L^{ - 1}}$, what is the value of $V$in mL, approximately?

Answer 1

Specific gravity and density are interchangeable terms, i.e. they have the same meaning in context of solutions. When extra solvent is added to the solution, the mass of solute remains the same, meaning that the moles of solute remain the same in the new solution. This can be equated to find the volume of extra solvent added to the solution.

Complete step by step answer:
We have been given that the specific gravity of the salt = 1.025. Also, $V$mL of water has been added to 1.0 L of this solution. So, the new volume of the solution is equal to the sum of $V$ and 100 mL of water.
${V_{new}} = (V + 1000)mL$
Also, the relation between specific gravity, mass and volume is given as:
$Mass = Density \times Volume$
Notice that density and specific gravity are interchangeable terms.
Since, no extra salt is added to the solution, the mass of salt in the solution remains the same and hence, the number of moles of salt remains the same in the new solution.
Equating the mass of salt in both solutions, we get:

$$\therefore (V + 1000) \times 1.02 = 1.025 \times 1000 \\\ \Rightarrow V = 4.9mL \approx 5mL \\\$$

**Thus, around 5 mL of water has been added to the solution.
**
Note: Definitions of several terms in the questions are important. The specific gravity is the ratio between the density of an object, and a reference substance. The specific gravity can tell us, based on its value, if the object will sink or float in our reference substance. Since, our reference object is water, density and specific gravity can be considered to be the same.