Question

10. 0.9631g of sodium hydroxide was dissolved to make a 100.0ml solution. 10.00ml of this solution was made to 250.0ml. What is the concentration of this new solution? 0.009631M 11. Exactly 0.337g of potassium permanganate were dissolved to make a 200.0ml solution. This solution was used in the titration with unknown iron (I) sulfate solution made by dissolving 1.00g-tablet iron sulfate to make 100.ml solution. An average of 21.10ml of KMnO, was required to titrate 25.00ml of the iron (I) Solution in the presence of sulfuric acid. What is the percentage of iron (II) sulfate in the tablet? 68.4% FVO

Answer 1

Given, Volume of the solution = 100.0 mL = 0.1 L (1 L = 1000 mL) Mass of sodium hydroxide (NaOH) = 0.9631 g Molar mass of sodium hydroxide (NaOH) = 39.997 g/mol molarity of NaOH = moles/liter Moles of NaOH = (mass/molar mass) = 0.9631 g/39.997 g/mol = 0.0240 mol Molarity = 0.0240 mol/0.1 L = 0.240 mol/L = 0.240 M For dilution M_1 V_1 = M_2 V_2 The NaOH solution is diluted from 10.0 mL to 250.0 mL. i.e., 25 times dilution (0.240 M/25 = 0.00963 M) or we can calculate from dilution formula M_1 = 0.240 M V_1 = 10.0 mL = 0.001 L M_2 = ? V_2 = 250 mL = 0.25 L M_2 = M_1 V_1 /V_2 = 0.240 M times 0.01 L/0.25 L = 0.0096317 = 9.63 times 10^-3 M Molarity of new NaOH solution = 9.63 times 10^-3 m