1. Problem: Find the molar concentration (M) of solutions containing 10 g/L of (a) NaOH, (b) Na₂SO₄, (c) K₂Cr₂O₇, and (d) KCl. Formula: Molarity $M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume of solution (L)}}$ (a) NaOH: Molar mass = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol $M = \frac{10}{40 \times 1} = 0.25$ M (b) Na₂SO₄: Molar mass = 2(23) + 32 + 4(16) = 142 g/mol $M = \frac{10}{142 \times 1} \approx 0.0704$ M (c) K₂Cr₂O₇: Molar mass = 2(39) + 2(52) + 7(16) = 294 g/mol $M = \frac{10}{294 \times 1} \approx 0.0340$ M (d) KCl: Molar mass = 39 (K) + 35.5 (Cl) = 74.5 g/mol $M = \frac{10}{74.5 \times 1} \approx 0.134$ M 2. Problem: Calculate the mass of potassium permanganate (KMnO₄) in 2 L of 0.15 M solution. Molar mass KMnO₄ = 39 (K) + 55 (Mn) + 4(16) = 158 g/mol Mass = Molarity $\times$ Volume $\times$ Molar mass $= 0.15 \times 2 \times 158 = 47.4$ g 3. Problem: Calculate (i) molal concentration, (ii) molar concentration, (iii) normal concentration, and (iv) mole fraction of 20.0% (w/w) ammonium sulphate solution with specific gravity 1.12. Given: 20 g solute in 100 g solution, density = 1.12 g/mL Mass solvent = 100 - 20 = 80 g = 0.08 kg Molar mass (NH₄)₂SO₄ = 2(14+4) + 32 + 4(16) = 132 g/mol (i) Molality $m = \frac{\text{moles solute}}{\text{kg solvent}} = \frac{20/132}{0.08} \approx 1.89$ mol/kg (ii) Volume of solution = $\frac{100}{1.12} = 89.29$ mL = 0.08929 L Molarity $M = \frac{20/132}{0.08929} \approx 1.70$ M (iii) Normality $N = M \times$ number of replaceable H ions = $1.70 \times 2 = 3.40$ N (iv) Mole fraction $X_{solute} = \frac{\text{moles solute}}{\text{moles solute} + \text{moles solvent}}$ Moles solvent (water) = $80/18 = 4.44$ mol $X_{solute} = \frac{20/132}{(20/132) + 4.44} \approx 0.033$ 4. Problem: A solution is prepared by dissolving 432 mg of potassium hexacyanoferrate (III) K₃Fe(CN)₆ in water and diluting to 1500 mL. Calculate: (a) Formal concentration of K₃Fe(CN)₆ Molar mass K₃Fe(CN)₆ = 3(39) + 55.8 + 6(12 + 14) = 329.2 g/mol Mass = 432 mg = 0.432 g Volume = 1.5 L Formal concentration = $\frac{0.432}{329.2 \times 1.5} \approx 0.000875$ M (b) Molar concentration of K⁺ Each formula unit has 3 K⁺ ions $= 3 \times 0.000875 = 0.00263$ M (c) Molar concentration of K₃Fe(CN)₆⁻ Same as formal concentration = 0.000875 M (d) Weight-volume percent of K₃Fe(CN)₆ $\% w/v = \frac{0.432}{1500} \times 100 = 0.0288\%$ (e) Number of millimoles of K⁺ in 100.00 mL Millimoles K⁺ = $0.00263 \times 100 = 0.263$ mmol Final answers: 1. (a) 0.25 M, (b) 0.0704 M, (c) 0.0340 M, (d) 0.134 M 2. Mass KMnO₄ = 47.4 g 3. (i) 1.89 mol/kg, (ii) 1.70 M, (iii) 3.40 N, (iv) 0.033 mole fraction 4. (a) 0.000875 M, (b) 0.00263 M, (c) 0.000875 M, (d) 0.0288% w/v, (e) 0.263 mmol