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# Units%20of%20Concentration%20Answers - f value of 1.86 C/m...

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CHEM 1212 Units of Concentration / Colligative Properties 1. For a solution that has 28 g of FePO 4 dissolved in water, the total mass of solution is 1.350 kg. What is the molality of this solution? 28 g of FePO 4 * ( 1 mol / 150.82 g) = 0.1857 mol m = 0.1857 mol / 1.322 kg = 0.14 mol/kg Note: mass of solvent was found by mass of solution – mass of solute 2. Assume the above solution has a density of 1.10 g/mL. What would be the molarity of the solution? The volume of the solution would be: 1350 g * ( 1 mL / 1.10 g) = 1227.3 mL = 1.227L M = 0.1857 mol / 1.227 L = 0.15 M 3. What would be the mole fraction for FePO 4 in the above solution? What would be the mole fraction for H 2 O in the above solution? Mol H 2 O = 1322 g * (1 mol / 18.02 g) = 73.36 mol χ FePO4 = 0.1857 mol / (0.1857 mol + 73.36 mol) = 0.0025 χ H2O = 73.36 mol / (0.1857 mol + 73.36 mol) = 0.9975 1.00 Note: the mole fractions of the two would add up to a value of 1.00 (or 100%) 4. With a k b value of 0.512 C/m and a k

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Unformatted text preview: f value of 1.86 C/m what would be the freezing point and boiling point of this solution? ∆T f = K f * m ∆T f = 1.86 * 0.14 * 2 ∆T f = 0.52 Freezing point = -0.52 Celsius ∆T b = K b * m ∆T b = 0.512 * 0.14 * 2 ∆T b = 0.14 Boiling point = 100.14 Celsius CHEM 1212 5. What is the osmotic pressure of the solution at 40.0 Celsius? Π = M * R * T Π = 0.15 M * 0.0821 L-atm/mol-K * T Π = 0.15 M * 2 * 0.0821 L-atm/mol-K * 313.15 Π = 7.71 atm 6. What is the vapor pressure of the solvent in this solution if the vapor pressure of pure water would be 92.5 torr at the temperature of the solution? P solvent = X solvent * P solvent X solvent = 73.36 mol / ( 2 * 0.1857 mol + 73.36 mol) = 0.9950 P solvent = 0.9950 * 92.5 torr P solvent = 92.0 torr Note, with vapor pressure, the “number of ions” factor, 2, goes with the moles of solute, which shows up in the denominator of the mole fraction....
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Units%20of%20Concentration%20Answers - f value of 1.86 C/m...

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