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 $1.02gm{L}^{-1}$, what is the value of $V$ in mL, approximately?

Answer 1

Given, the specific gravity = $1.025$
Also, $V$ mL of water has been added to $1.0$ L of this solution.
$V_{New}=(V+1000)$ mL
Number of moles of salt remains the same.
$\therefore (V+1000)\times 2.02=1.025\times 1000$
$V=4.9$ mL $\approx 5$ mL
Thus, around $5$ mL of water has been added.