Solutions — Questions & Answers

Sample questions from Solutions

1. Q1. For an aqueous solution of a nonelectrolyte, which pair of measured values is mutually INCONSISTENT? (K_f = 1.86, K_b = 0.52 K kg/mol)

   • A.Delta_T_f = 0.372 K and Delta_T_b = 0.104 K
   • B.Delta_T_f = 0.930 K and Delta_T_b = 0.260 K
   • C.Delta_T_f = 1.860 K and Delta_T_b = 0.520 K
   • D.Delta_T_f = 0.372 K and Delta_T_b = 0.520 K✓
   Solution

   For the same solution the molality is the same for both measurements, so Delta_T_f/Delta_T_b must equal K_f/K_b = 1.86/0.52 = 3.58. Checking: A gives 0.372/0.104 = 3.58 (consistent); B gives 0.930/0.260 = 3.58 (consistent); C gives 1.860/0.520 = 3.58 (consistent); D gives 0.372/0.520 = 0.72, which is far from 3.58 and therefore impossible for a single solution.

2. Q2. A solute is found to show ABNORMAL MOLAR MASS when determined via a colligative property measurement (i.e. the molar mass calculated from the colligative property does not match the solute's expected/true molar mass). Which situation would cause the CALCULATED molar mass to appear LOWER than the true molar mass?

   • A.Abnormal molar mass can never occur for any real solute under any circumstances
   • B.The solute undergoes ASSOCIATION in solution (multiple solute molecules combining together, e.g. via hydrogen bonding, to form fewer, larger effective particles)
   • C.The solute undergoes DISSOCIATION in solution (breaking apart to multiple ions/particles, e.g. an ionic compound like NaCl dissociating to Na+ and Cl- ions), increasing the effective number of particles present beyond what the true molar mass alone would suggest✓
   • D.The solute's true, unmodified molar mass always exactly matches the colligative-property-calculated value, with no exceptions
   Solution

   Colligative properties depend on the NUMBER of solute PARTICLES present, and the standard formulas (relating colligative property magnitude to molality/concentration) implicitly assume each solute FORMULA UNIT remains as ONE single particle in solution -- if the solute instead DISSOCIATES to multiple particles (e.g. an ionic compound like NaCl dissociating to separate Na+ and Cl- ions), the ACTUAL number of particles present is GREATER than the standard formula assumes (based on the compound's nominal molar mass) -- since colligative properties scale with particle NUMBER, this greater-than-expected particle count produces a LARGER observed colligative effect than the true molar mass would predict, and working BACKWARD from this larger observed effect (using the standard formula) yields a CALCULATED molar mass LOWER than the true molar mass -- the reverse (ASSOCIATION, where multiple molecules combine to fewer, larger effective particles) would instead produce a calculated molar mass HIGHER than the true value.

3. Q3. The partial pressure of a component in an ideal solution is plotted against its mole fraction in the liquid. For the same pair of liquids, which pair of statements about the slope and intercept is correct?

   • A.Slope = p^0 of the other component; intercept at x = 0 equals p^0 of that component
   • B.Slope = p^0 of that component; intercept at x = 0 is zero✓
   • C.Slope is zero; intercept equals the total vapour pressure
   • D.Slope = K_H of the component; intercept at x = 0 equals p^0
   Solution

   Raoult law for a component of an ideal solution reads p_A = p_A^0 x_A, which is a straight line through the ORIGIN whose slope is the vapour pressure of PURE A. At x_A = 0 there is no A present, so its partial pressure is zero, giving a zero intercept. It is the TOTAL vapour pressure line that has the nonzero intercept p_B^0 at x_A = 0.

4. Q4. The ELEVATION OF BOILING POINT (delta-T_b) of a solution is given by delta-T_b = K_b x m (where K_b is the molal elevation constant/ebullioscopic constant, and m is the molality of the solute). Why does dissolving a nonvolatile solute RAISE a solvent's boiling point?

   • A.Boiling point elevation occurs because the solute physically blocks the solvent from evaporating at the molecular level, unrelated to vapour pressure
   • B.Dissolving a solute always increases the solvent's vapour pressure, directly raising the boiling point
   • C.Since the solution's vapour pressure is LOWERED (due to the dissolved solute, as described by Raoult's law), a HIGHER temperature is needed for the solution's vapour pressure to reach atmospheric pressure (the condition for boiling), hence the boiling point is raised✓
   • D.Boiling point elevation only occurs for solid solutes, never for liquid solutes
   Solution

   A liquid BOILS when its vapour pressure equals the surrounding atmospheric pressure -- since dissolving a nonvolatile solute LOWERS the solution's vapour pressure (compared to the pure solvent, at any given temperature, per Raoult's law and the relative-lowering-of-vapour-pressure relationship), the solution's vapour pressure will only reach atmospheric pressure (the boiling condition) at a HIGHER temperature than the pure solvent requires -- this is precisely why dissolving a nonvolatile solute RAISES (elevates) a solvent's boiling point, and why this boiling-point elevation, like vapour-pressure lowering, is a colligative property, depending on the solute's molal concentration (particle number per kg of solvent) rather than its specific chemical identity.

5. Q5. A solution shows POSITIVE deviation from Raoult's law. A student argues that, since positive deviation means the solution's vapour pressure is HIGHER than ideal, this solution should therefore have a LOWER boiling point than a corresponding ideal solution of the same composition. Evaluate this argument.

   • A.The argument is CORRECT -- since a HIGHER vapour pressure at any given temperature means the vapour pressure reaches atmospheric pressure at a LOWER temperature, a solution with positive deviation (higher-than-ideal vapour pressure) boils at a LOWER temperature than an ideal solution of the same composition would, consistent with minimum-boiling azeotrope formation for strong positive deviation✓
   • B.The argument is incorrect because positive deviation always leads to a HIGHER boiling point, never a lower one
   • C.The argument is incorrect because vapour pressure and boiling point are entirely unrelated quantities
   • D.The argument is incorrect because positive deviation has no consistent effect on boiling point in either direction
   Solution

   The student's argument is CORRECT -- POSITIVE deviation from Raoult's law means the ACTUAL vapour pressure of the solution, at any given temperature, is HIGHER than the vapour pressure predicted by ideal (Raoult's-law) behaviour for the same composition -- since a liquid boils when its vapour pressure equals the surrounding atmospheric pressure, a solution with a HIGHER vapour pressure (at a given temperature) will reach the boiling condition (atmospheric pressure) at a LOWER temperature than an ideal solution of the same composition would require -- therefore, positive deviation is correctly associated with a LOWER boiling point (relative to ideal behaviour), and, for sufficiently strong positive deviation, a MINIMUM-BOILING AZEOTROPE forms at the specific composition where this vapour-pressure-raising (and hence boiling-point-lowering) effect is maximal -- this is the OPPOSITE of the maximum-boiling-azeotrope situation that arises from strong NEGATIVE deviation, and the classic example of a minimum-boiling azeotrope is the ethanol-water system, which is why ethanol cannot be purified beyond approximately 95-96% by simple fractional distillation alone.

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