Question

Find the volume of water should be added to 50 mL of $HN{O}_3$ having density $1.5gm{L}^{-1}$ and 63.0% by weight to have one molar solution. (Give your answer in mL)

Answer 1

Volume of $HN{O}_3=50mL,$
density = 1.5 g/mL
$density=\frac{M}{V},$
mass of given $HN{O}_3$ $=50\times 1.5=75g$
it is $63\%$ by weight so,
actual weight of $HN{O}_3=\frac{75\times 63}{100}=\frac{3}{4}\times 63$

Moles of $HN{O}_3=\frac{3}{4}\times \frac{63}{63}=\frac{3}{4}$

therefore using molarity of solution we have:
$M=\frac{MolesofHN{O}_3}{Volumeofsolution}=\frac{3}{4\times V_L}=1$
volume of solution $=\frac{3}{4}L=750mL$
so, Volume of water required = 750 - 50 = 700 mL