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13 Properties of Solutions Solutions to Exercises (c) The molecules have similar molar masses. NO is most soluble because it is polar. The triple bond in N₂ is shorter than the double bond in It is more difficult for molecules to surround the smaller N₂ molecules, so they are less soluble than O₂ molecules. (d) and are polar molecules capable of hydrogen bonding with water. Hydrogen bonding is the strongest force between neutral molecules and causes the much greater solubility. is weakly acidic in water. reacts with water to form a weak acid. The large solubility of SO₂ is a sure sign that a chemical process has occurred. (e) N₂ and N₂ is too small to be easily hydrated, so is more soluble in NO (31) and (30). The structures of these two molecules are very different, yet they have similar solubilities. NO is slightly polar, but too small to be easily hydrated. The larger is nonpolar, but more polarizable (stronger dispersion forces). NO (31) and O₂ (32). The slightly polar NO is more soluble than the slightly larger (longer 0=0 bond than N = bond) but nonpolar O₂. 13.112 The resulting solution is very dilute, so assume ideal behavior. Assume the amount of water consumed in the reaction is negligible. Ignore the solubility of H₂(g) in the solution (see Solution 3.110). 1.0 mm 3 0.535 cm³ X 10³ 1³ cm³ 3 = 5.35 10⁻⁴ = 5.4 10⁻⁴ g Li 5.35 X 10⁻⁴ g Lix 6.941 1 mol Li = 7.708 X 10⁻⁵ = 7.7 10⁻⁵ mol Li mol Li = mol LiOH; 2 mol ions per mol LiOH 7.708 X 10⁻⁵ mol Li = 7.708 10⁻⁵ mol LiOH = 1.542 X 10⁻⁴ mol ions = 1.5 X 10⁻⁴ mol ions m = 1.542 0.500L 10⁻⁴ mol ions 1000 1L mL 0.997 1mL H₂O 1000 1kg = 3.092 = 10⁻⁴ m = m = -1.86(3.092 = -5.8 °C; Tf = 0.00000 0.00058 = -0.00058 °C The freezing point of the LiOH(aq) solution is essentially zero. 13.113 (a) For an ideal solution, Raoult's Law is obeyed. Pₜ = + = 0.5(300 torr) = 150 torr = 0.5(360 torr) = 180 torr; Pₜ = 150 torr + 180 torr = 330 torr 393