MCAT Non-Sim General Chemistry GCHM - Solutions 5 years Reyes, V. M. Question Information Q-Bank MCAT Sim Non-Sim Subject General Chemistry Foundation GCHM - Solutions Validity 5 years Author(s) Reyes, V. M. Reviewer(s) 0000000 Editor(s) 0000000 Passage Information Passage ID Media ID(s) Passage Stem The word “colligative” derives from the word “co-ligate”, with “co” meaning “together” and “ligate” meaning “to bind” or “be bound”. Thus “colligative” implies a “binding together”, and in the case of chemical solutions, between solvent and solute molecules. Colligative property is only applicable to liquid or suitably fluid solutions and not to solid solutions (as in alloys) because in solids, molecules possess no Brownian motion upon which colligative properties depend. Also, since most liquid solutions encountered in chemistry are those where water is the solvent, we shall limit our discussion here to aqueous solutions. Colligative properties of a solution of a particular solute have the effect of raising the boiling point of the solvent, depressing its freezing point, and lowering both its vapor pressure and osmotic pressure. We shall limit our discussion to boiling point elevation and freezing point depression in this passage.

Consider a liter of water where a few grams of the salt KCl is dissolved. Since KCl is an ionic compound, each molecule of it dissociates in water into two ion particles, K+ and Cl- (see (1)). The so-called “van’t Hoff factor,” a unitless number, for KCl is thus 2. KCl K+ + Cl- (1) Next consider a liter of water where a few grams of the salt CaCl2 is dissolved. CaCl2 is also an ionic compound, and it can be assumed that for every molecule of CaCl2 an ion of Ca++ and two ions of Cl- are produced (see (2)), yielding three ion particles. Its van’t Hoff factor is thus 3. CaCl2 Ca++ + 2Cl- (2) Finally consider a liter of water where a few grams of table sugar, sucrose (C12H22O11), is dissolved. Sucrose is a covalent dimer consisting of fructose and glucose. It is not an ionic compound but a covalent one. It does not dissociate into its constituent monomers, glucose and fructose, which are covalently bound. When dissolved in water; sucrose remains a dimer (see (3)). Thus its van’t Hoff factor is 1. C12H22O11 C12H22O11 (no dissociation) (3) The van’t Hoff factor of the solute affects the colligative properties of solutions (assumed aqueous here) of that solute. Colligative properties are properties of a solution V made up of solvent S and of solute t that alters the boiling point, freezing point, vapor pressure and osmotic pressure of V relative to those of pure solvent S. Colligative properties are functions of the number of solute particles t dissolved in S – and thus of the soolute’s van’t Hoff factor - but not at all by the identity or nature of the solute t molecules. We shall focus here on the boiling point (BP) elevation and freezing point (FP) depression of a solution V composed of solvent S and solute t. We must first know the concentration of solute t in solution V in terms of molality, which is the number of moles of t dissolved per kilogram of S; let’s assume it has value m. Then the BP elevation and FP depression are given respectively by:

Maron, S.H. & Prutton, C.F., Physical Chemistry In (4), ∆Tb is the BP elevation and Kb is the ebullioscopic constant (a.k.a. BP constant), of solvent S. Similarly, in (5), ∆Tf is the FP depression and Kf is the cryoscopic constant (a.k.a. FP constant) of S. In both (4) and (5), mt is the molality of solute t, and it is its van’t Hoff factor (vHf). More on the van’t Hoff factor: In our discussions of the vHf regarding the solute dissociations in (1), (2) and (3) above, we assumed that the solutions behave ideally, i.e., that the dissociations are complete, i.e., 100%. In reality this is attained only if the solutions are extremely dilute. As the solution become more concentrated, the dissociation becomes less complete because there is a greater chance of reassociation between the components of the solute. Thus in practical terms, the vHf in (1) is slightly less than 2, that in (2) is slightly less than 3, and that in (3) is slightly less than 1. More on boiling point (as well as vapor pressure): Since BP depends on the ambient air pressure, and ambient air pressure is dependent on the altitude, BP (and vapor pressure) also depends on altitude. This is why it takes a shorter time for water to boil in Mexico City than in Amsterdam because the former is over 2,240 meters above sea level, while the later is 2 meters below sea level. The boiling point values given in handbooks of physics or chemistry are usually at sea level. For water, the BP at sea level is 100.0 ᵒC. Freezing point, on the other hand, is not affected by altitude because it is a liquid-solid phase transition and thus independent of the ambient air pressure. Passage References PMID/Book Title of Publication or Book 000000000 Maron, S.H. & Prutton, C.F., Physical Chemistry 000000000 Daniels, F. & Alberty, R., Experimental Physical Chemistry (N/A) en.wikipedia.com, youtube.com (N/A) Author’s own lecture notes.

Question Attributes #1 (QID: 000000) Topic Blueprint Colligative Properties Competency MCAT: BS-2: Application of Concepts & Principles To understand the colligative properties of solutions, and be able to calculate how much a given amount of solute affects the solvent’s boiling point and freezing point. Objective Media ID(s) 00000000 Question ID 00000000 Question Stem #1 In 1 kg. of pure water is dissolved 8.713 grams of potassium sulfate. Determine the boiling and freezing points of the resulting solution. The MW of potassium sulfate is 174.259 g/mol. The ebullioscopic and cryoscopic constants of water are, respectively, 0.513 ᵒC kgsolvent per molesolute and 1.858 ᵒC kgsolvent per molesolute . Assume an elevation of sea level. (cont’d. on following page)

molality of the resulting solution is 0.050 m. Answer Choices #1 A) BP = 100.077 ᵒC & FP = 0.279 ᵒC B) BP = 100.077 ᵒC & FP = -0.279 ᵒC C) BP = 100.051 ᵒC & FP = -0.186 ᵒC D) BP = 99.721 ᵒC & FP = 0.077 ᵒC Correct: B) Explanation #1 The correct answer is B. Potassium sulfate is K2SO4 and it is freely soluble in water as: K2SO4 2K+ + SO4= We thus assume a van’t Hoff factor of 3. The number of moles of K2SO4 dissolved in 1 kg. of water is (8.713 gms.)/(174.259 gms./mol) = 0.050 moles, thus the molality of the resulting solution is 0.050 m. The BP elevation is 100.0 ᵒC + (0.513)(0.050)(3) = 100.077 ᵒC The FP depression is 0.0 ᵒC - (1.858)(0.050)(3) = -0.279 ᵒC (Choice A) This choice is incorrect because the FP depression was added to, instead of subtracted from, the FP of pure water. (Choice C) This choice is incorrect because the van’t Hoff factor used was 2 instead of 3 (one K+ and two SO4= ions) (Choice D) This choice is incorrect because the BP elevation was subtracted from, instead added to, the BP of pure water; the FP depression was also added to, instead of subtracted from, the FP of pure water.

point and freezing point. Educational objective: To understand the colligative properties of solutions, and be able to calculate how much a given amount of solute affects the solvent’s boiling point and freezing point. References #1 PMID/Book Title of Publication or Book 0000000 000000000 000000000 0000000 Verifications #1 Yes / No The question is at the Application or higher cognitive level. Yes / No The question is based on a realistic clinical scenario. Yes / No The question has at least one close distracter, and other options have educational value. Yes / No The question is appropriate to the entry level of nursing practice. Yes / No The explanation is short and concise, yet thorough. Yes / No The question has an appropriate table/flow chart/illustration.