Question

Calculate molarity and molality of 6.3% of solution of nitric acid having density 1.04$gc{m}^{-3}$.

($H=1,N=14,O=16$)

Answer 1

Density of nitric acid solution $=1.04gmc{m}^{-3}$

Volume of $100gm$ solution $=\frac{100}{1.04}=96.15c{m}^3=96.15\times {10}^{-3}d{m}^3$

Molarity $=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution in L}}$

$=\frac{6.3}{63\times 96.15\times {10}^{-3}}=1.04mol/L$

$6.3$% $HN{O}_3$ means $6.3g$ of $HN{O}_3$ present in $100gm$ of solution

$\therefore$ Mass of water $=100-6.3=93.7g$

Molality of $HN{O}_3=\frac{\text{Mass of}HN{O}_3}{\text{Molar mass of}HN{O}_3\times \text{Mass of solvent}\left(kg\right)}$

$=\frac{6.3}{63\times 93.7\times {10}^{-3}}$

$=1.067mol/Kg$