Question

Calculate molarity and molality of 6.3% solution of nitric acid having density ${1.04gcm}^{-3}$($H=1$, $N=14$, $O=16$)

Answer 1

Given density of $6.3$% $HN{O}_3=1.04g/c{m}^3$, Molar mass = $63g$

(a) Molarity= $\frac{\text{Number of moles of solute}}{\text{Volume of solution}}$

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

$\therefore$ Molarity= $\frac{6.3\times {10}^{-3}}{63\times {10}^{-3}\times 96.15\times {10}^{-3}}=1.04mol/d{m}^3$

(b) Molality= $\frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}$

Mass of $H_{2}O=100-6.3=93.7g$

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

$=1.067mol/kg$ .