Molarity vs. molality - 2 | Chemical Calculations

Solution 1: 30.0 mL of 6.0M HCl

Solution 2: 470 mL of 2.0M HCl

Solution 3: final 500-mL solution

The number of moles in solution does not change:

n₁ + n₂ = n₃

⇒ M₁ V₁ + M₂ V₂ = M₃ V₃

⇒ M₃ = $\frac{M_1 {\mathrm{V}}_1 + M_2 {\mathrm{V}}_2}{{\mathrm{V}}_3}$

⇒ M₃ = $\frac{6.0 \times 30.0 + 2.0 \times 470}{500}$ = 2.24 mol.L^-1