Question

Calculate molarity and molality of 6.3% solution of nitric acid having density 1.04 g cm⁻³
. (H = 1, N = 14, O = 16)

Answer 1

Given: Density of 6.3% HNO₃ = 1.04 g cm⁻³

To find: a. Molarity b. Molality

Formulae:

a) Molarity = Number of moles of solute/Volume of solution in L b) Molality = Number of moles of solute/Mass of solvent in kg

Calculation: Density of solution = Mass of solution/Volume of solution

∴ Volume of solution = ${\displaystyle \frac{100}{1.04}}$ = 96.15 cm³ = 96.15 × 10^–3 dm³

Molarity = ${\displaystyle \frac{\text{Massof solute}}{\text{Molar mass of solute×Volume of solution in L}}}$

∴ Molarity  = ${\displaystyle \frac{6.3\times {10}^{-3}}{63\times {10}^{-3}\times 96.15\times {10}^{-3}}}$ = 1.04 mol/dm³

b) 6.3% HNO3 is present means 6.3 g of HNO3 is present in 100 g of solution.

Mass of H₂O = 100 − 6.3 = 93.7 g

Molality of HNO₃ = Massof HNO₃/Molar mass of HNO₃ ×Massof solvent in kg

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

= 1.067 mol/kg