1. **Problem statement:** A 0.15 molal solution of carbonic acid (H2CO3) is prepared. We need to find: a. Molarity b. Normality c. Mole fraction of solute and solvent d. Volume-volume percent of solute to solvent (density of acid = 1.49 g/mL) e. Concentration in ppm 2. **Given data:** - Molality, $m = 0.15$ mol/kg solvent - Density of carbonic acid, $\rho = 1.49$ g/mL - Molar mass of carbonic acid, $M = 62.03$ g/mol - Assume 1 kg (1000 g) of solvent (water) for calculation --- ### a. Molarity ($M$): Molarity is moles of solute per liter of solution. - Moles of solute = $0.15$ mol (since molality is moles per kg solvent and solvent = 1 kg) - Mass of solute = $0.15 \times 62.03 = 9.3045$ g - Mass of solution = mass solute + mass solvent = $9.3045 + 1000 = 1009.3045$ g - Volume of solution = $\frac{\text{mass of solution}}{\text{density of solution}}$. Assuming density of solution approximately equals density of water (since dilute), $\approx 1$ g/mL = 1000 g/L. Volume of solution $\approx \frac{1009.3045}{1000} = 1.0093$ L Molarity $= \frac{0.15}{1.0093} = 0.1486$ mol/L --- ### b. Normality ($N$): Carbonic acid is diprotic (2 acidic protons), so normality = molarity $\times$ number of equivalents. $N = 0.1486 \times 2 = 0.2972$ eq/L --- ### c. Mole fraction ($X$): - Moles of solvent (water) = $\frac{1000}{18.015} = 55.51$ mol - Moles of solute = 0.15 mol Mole fraction of solute: $$X_{solute} = \frac{0.15}{0.15 + 55.51} = 0.0027$$ Mole fraction of solvent: $$X_{solvent} = \frac{55.51}{0.15 + 55.51} = 0.9973$$ --- ### d. Volume-volume percent of solute to solvent: - Volume of solute = $\frac{\text{mass}}{\text{density}} = \frac{9.3045}{1.49} = 6.245$ mL - Volume of solvent = 1000 mL (1 kg water) Volume-volume percent: $$\% = \frac{6.245}{1000} \times 100 = 0.6245\%$$ --- ### e. Concentration in ppm: - ppm = mg solute per kg solution - Mass of solute = 9.3045 g = 9304.5 mg - Mass of solution = 1009.3045 g = 1.0093 kg $$\text{ppm} = \frac{9304.5}{1.0093} = 9220.5$$ ppm --- **Final answers:** - a. Molarity = $0.1486$ mol/L - b. Normality = $0.2972$ eq/L - c. Mole fraction solute = $0.0027$, solvent = $0.9973$ - d. Volume-volume percent = $0.6245\%$ - e. Concentration = $9220.5$ ppm