Question

Find the volume of water (in mL) added to 50 mL concentrated nitric acid $HN{O}_3$
$(d=1.52gm{l}^{–1}$, 69% by mass) that forms diluted nitric acid ($d=1.15gm{L}^{–1}$, 19% by mass $HN{O}_3$).

Answer 1

Let volume of dilute solution that can be formed is V.
Mass of $HN{O}_3$ will be same before and after dilution
So, $V\times 1.15\times 0.19=50\times 1.52\times 0.69$
$⟹V=240$ mL
Volume of water added = 240 - 50 = 190 mL.