1. **State the problem:** We need to prepare a 0.5 normal (0.5 N) sulfuric acid solution using concentrated sulfuric acid with 96% purity and specific gravity 1.84. 2. **Formula and concepts:** Normality (N) relates to equivalents per liter. For sulfuric acid (H2SO4), 1 mole provides 2 equivalents because it can donate 2 H+ ions. 3. **Calculate the molarity of the concentrated acid:** - Specific gravity (SG) = 1.84 means 1 L weighs 1.84 kg = 1840 g. - Purity = 96%, so mass of pure H2SO4 in 1 L = 1840 g × 0.96 = 1766.4 g. - Molar mass of H2SO4 = 98 g/mol. - Moles in 1 L = \frac{1766.4}{98} \approx 18.02 \text{ mol}. 4. **Calculate normality of concentrated acid:** - Normality = Molarity × equivalents per mole = 18.02 × 2 = 36.04 N. 5. **Use dilution formula to find volume of concentrated acid needed:** $$N_1 V_1 = N_2 V_2$$ Where: - $N_1 = 36.04$ N (concentrated acid normality) - $V_1 = ?$ (volume of concentrated acid needed) - $N_2 = 0.5$ N (desired normality) - $V_2 = 1$ L (final volume of solution) 6. **Solve for $V_1$:** $$V_1 = \frac{N_2 V_2}{N_1} = \frac{0.5 \times 1}{36.04} \approx 0.01388 \text{ L} = 13.88 \text{ mL}$$ 7. **Interpretation:** To prepare 1 L of 0.5 N sulfuric acid, measure 13.88 mL of the concentrated acid and dilute with distilled water up to 1 L. **Final answer:** Use 13.88 mL of concentrated sulfuric acid (96% purity, SG 1.84) and dilute to 1 L to get 0.5 N solution.