Question

240 ml of diluted nitric acid (d = 1.15 g/ml, 19% by mass) $HN{O}_3$ is to be prepared from conc. $HN{O}_3$ (d =1.52 g/ml, 69% by mass). How much water has to be added (in mL) to the concentrated solution to achieve this?

• A. 160 mL
• B. 190 mL
• C. 50 mL
• D. 30 mL

Answer 1

The correct option is B 190 mL

The mass of the acid before and after dilution will be the same, the water added changes the density and also the concentration (mass % - w/w).
So, the mass of the acid before dilution = mass after dilution
Mass of the acid: $\text{density}\times$ $\frac{\text{mass \%}}{100}\times \text{total volume}$
If $V$ is the volume of water added in mL to concentrated acid to make it up to 240 mL of diluted acid,
Mass before dilution: $1.52\times 0.69\times (240-V)$
Mass after dilution: $1.15\times 0.19\times 240$
$V=[1-(\frac{1.15}{1.52}\times \frac{0.19}{0.69}\left)\right]$$\times 240=190$ mL