[{"data":1,"prerenderedAt":196},["ShallowReactive",2],{"topic-colligative-properties":3},{"topic":4,"prev":180,"next":188},{"id":5,"slug":6,"title":7,"shortTitle":7,"lesson":5,"category":8,"description":9,"metaDescription":10,"objectives":11,"conceptSections":27,"workedExample":93,"oerSources":103,"selfStudyQuestions":105,"relatedTopicSlugs":125,"relatedElements":128,"relatedReferences":133,"keyEquations":135,"howTo":149,"conceptProbe":157,"practiceTeaser":161,"gatedContent":169,"seoKeywords":172,"lastAlignmentAudit":178,"objectiveCount":179},19,"colligative-properties","Colligative Properties","states-of-matter","Understand colligative properties: boiling point elevation, freezing point depression, osmotic pressure, and Raoult's law.","Learn colligative properties: boiling point elevation, freezing point depression, osmotic pressure, and Raoult's law. Van't Hoff factor for electrolytes.",[12,15,18,21,24],{"id":13,"text":14},"19.1","Calculate boiling-point elevation and freezing-point depression for a solution",{"id":16,"text":17},"19.2","Calculate osmotic pressure of a solution using π = iMRT",{"id":19,"text":20},"19.3","Apply Raoult's law to calculate the vapor-pressure lowering of a solution",{"id":22,"text":23},"19.4","Use the van 't Hoff factor (i) to account for dissociation of electrolytes in colligative-property calculations",{"id":25,"text":26},"19.5","Determine the molar mass of a solute from colligative-property measurements",[28,38,47,55,63,72,81,89],{"heading":29,"content":30,"relatedObjectives":31,"deepDive":34},"What Are Colligative Properties?","\u003Cp>\u003Cstrong>Colligative properties\u003C/strong> are physical properties of solutions that depend only on the \u003Cem>number\u003C/em> of dissolved solute particles, not on their chemical identity. The four colligative properties are vapour-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure.\u003C/p>\u003Cp>The underlying cause is the same for all four: dissolved solute particles reduce the tendency of solvent molecules to escape into the vapour phase. Because these effects depend on particle count, electrolytes that dissociate into multiple ions produce a larger colligative effect than the same molality of a nonelectrolyte.\u003C/p>\u003Cp>Each colligative-property equation uses a different concentration unit:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Boiling-point elevation\u003C/strong> and \u003Cstrong>freezing-point depression\u003C/strong> use \u003Cstrong>molality (m)\u003C/strong> — moles of solute per kilogram of solvent. Molality is independent of temperature, which makes it the natural choice when the solution is being heated to its boiling point or cooled to its freezing point.\u003C/li>\u003Cli>\u003Cstrong>Osmotic pressure\u003C/strong> uses \u003Cstrong>molarity (M)\u003C/strong> — moles of solute per liter of solution — because the equation Π = iMRT is derived from the same gas-law-style framework that uses volume.\u003C/li>\u003Cli>\u003Cstrong>Raoult’s law\u003C/strong> uses \u003Cstrong>mole fraction\u003C/strong> (χ) — the fraction of total moles that the solvent represents.\u003C/li>\u003C/ul>\u003Cp>Converting between molarity, molality, mass percent, and mole fraction (using density to bridge volume and mass) is the subject of T9 LO 9.3 — see \u003Cstrong>\u003Ca href=\"/learn/solutions-and-concentration\">T9: Solutions and Concentration\u003C/a>\u003C/strong> for the four-way conversion workflow. This topic assumes you arrive with the right concentration unit for the equation you’re using.\u003C/p>\u003Cp>\u003Cstrong>Molality of a saturated solution.\u003C/strong> Solubility data are usually reported as grams of solute per 100 g of water at a given temperature. Convert that directly to molality by computing moles of solute (mass ÷ molar mass) and dividing by the mass of water in kilograms. Without a measured density of the saturated solution, molarity cannot be inferred from solubility data alone — molality is the natural concentration unit for these problems.\u003C/p>\u003Cp>\u003Cstrong>Effective particle concentration.\u003C/strong> When a strong electrolyte dissolves, the relevant particle count is the sum of every ion released, not the formula-unit count. Each ion concentration equals the formula-unit concentration multiplied by the subscript of that ion in the formula. For 0.250 M CaCl\u003Csub>2\u003C/sub>: [Ca\u003Csup>2+\u003C/sup>] = 0.250 M, [Cl\u003Csup>−\u003C/sup>] = 2 × 0.250 = 0.500 M, total dissolved-particle concentration = 0.750 M. For 0.500 m Al\u003Csub>2\u003C/sub>(SO\u003Csub>4\u003C/sub>)\u003Csub>3\u003C/sub>: [Al\u003Csup>3+\u003C/sup>] = 1.00 m, [SO\u003Csub>4\u003C/sub>\u003Csup>2−\u003C/sup>] = 1.50 m, total = 2.50 m. This total is the input to every colligative-property equation below, encapsulated by the van’t Hoff factor i.\u003C/p>",[32,33],145,146,[35],{"label":36,"body":37},"Why this matters: road salt and radiator fluid are particle counters","\u003Cp>Winter road crews and your car’s cooling system both exploit the central fact of this topic: only the \u003Cem>number\u003C/em> of dissolved particles matters, not what they are. Salt on an icy road lowers water’s freezing point below the air temperature; ethylene glycol in a radiator works both directions at once, keeping coolant liquid on the coldest morning (freezing-point depression) and unboiled on the hottest climb (boiling-point elevation). Neither substance reacts with the water; each simply shows up in enormous particle numbers. Colligative chemistry is crowd behavior: the solvent responds to how many guests arrived, never to who they are.\u003C/p>",{"heading":39,"content":40,"relatedObjectives":41,"deepDive":43},"Raoult's Law and Vapour-Pressure Lowering","\u003Cp>\u003Cstrong>Raoult’s law\u003C/strong> states that the vapour pressure of a solvent above a solution equals the mole fraction of the solvent times the vapour pressure of the pure solvent:\u003C/p>\u003Cp>\u003Cstrong>P\u003Csub>solution\u003C/sub> = χ\u003Csub>solvent\u003C/sub> · P°\u003Csub>solvent\u003C/sub>\u003C/strong>\u003C/p>\u003Cul>\u003Cli>χ\u003Csub>solvent\u003C/sub> = mole fraction of the solvent = mol solvent / total mol\u003C/li>\u003Cli>P°\u003Csub>solvent\u003C/sub> = vapour pressure of the pure solvent at that temperature\u003C/li>\u003C/ul>\u003Cp>Because adding a nonvolatile solute reduces χ\u003Csub>solvent\u003C/sub> below 1, the solution’s vapour pressure is always \u003Cem>lower\u003C/em> than that of the pure solvent. The magnitude of the lowering is ΔP = χ\u003Csub>solute\u003C/sub> · P°\u003Csub>solvent\u003C/sub>.\u003C/p>\u003Cp>This vapour-pressure lowering is the root cause of boiling-point elevation and freezing-point depression — all colligative properties trace back to the reduced tendency of solvent molecules to escape into the gas phase.\u003C/p>",[42],148,[44],{"label":45,"body":46},"Why dissolved particles lower vapour pressure","\u003Cp>The molecular picture behind Raoult’s law is disarmingly simple: evaporation happens only at the surface, and a nonvolatile solute dilutes that surface. Some fraction of the top layer is now occupied by particles that cannot leave, so fewer solvent molecules per second escape into the vapour, while the return traffic (condensation) is unchanged. The equilibrium settles at a lower vapour pressure, in direct proportion to how much of the liquid is still actually solvent: exactly the mole-fraction form of the law.\u003C/p>\u003Cp>Every other colligative property on this page is downstream of this one surface-crowding effect.\u003C/p>",{"heading":48,"content":49,"relatedObjectives":50,"deepDive":51},"Boiling-Point Elevation","\u003Cp>Adding a nonvolatile solute to a solvent \u003Cem>raises\u003C/em> its boiling point. Because the solute lowers the vapour pressure, a higher temperature is needed for the vapour pressure to reach atmospheric pressure (the condition for boiling).\u003C/p>\u003Cp>The equation: \u003Cstrong>ΔT\u003Csub>b\u003C/sub> = i · K\u003Csub>b\u003C/sub> · m\u003C/strong>\u003C/p>\u003Cul>\u003Cli>ΔT\u003Csub>b\u003C/sub> = boiling-point elevation (°C)\u003C/li>\u003Cli>i = van’t Hoff factor (number of particles per formula unit)\u003C/li>\u003Cli>K\u003Csub>b\u003C/sub> = ebullioscopic constant of the solvent (for water, 0.512 °C/m)\u003C/li>\u003Cli>m = molality (mol solute / kg solvent)\u003C/li>\u003C/ul>\u003Cp>The new boiling point = normal boiling point + ΔT\u003Csub>b\u003C/sub>. For example, a 1.00 m aqueous glucose solution (i = 1) boils at 100.00 + (1)(0.512)(1.00) = 100.51 °C. The small size of K\u003Csub>b\u003C/sub> means boiling-point elevation is a relatively subtle effect in dilute solutions.\u003C/p>",[32],[52],{"label":53,"body":54},"Common mistake: salting pasta water to cook faster","\u003Cp>The kitchen legend says salted water boils hotter and cooks pasta faster. Run the numbers: a generous tablespoon of salt (about 18 g, or 0.31 mol NaCl) in 4 kg of water gives a molality of 0.078 m, and with i = 2 for NaCl, ΔT\u003Csub>b\u003C/sub> = i·K\u003Csub>b\u003C/sub>·m = 2 × 0.512 × 0.078 ≈ 0.08 °C: under a tenth of a degree. Your pasta cannot tell.\u003C/p>\u003Cp>Salt the water for flavor, which works, not for physics, which does not at kitchen concentrations. Boiling-point elevation is real; the mistake is expecting a colligative effect without a colligative-scale particle count.\u003C/p>",{"heading":56,"content":57,"relatedObjectives":58,"deepDive":59},"Freezing-Point Depression","\u003Cp>Adding a solute \u003Cem>lowers\u003C/em> the freezing point of a solution. Solute particles disrupt the formation of the ordered crystal lattice that defines the solid phase, so a lower temperature is required for the solvent to freeze.\u003C/p>\u003Cp>The equation: \u003Cstrong>ΔT\u003Csub>f\u003C/sub> = i · K\u003Csub>f\u003C/sub> · m\u003C/strong>\u003C/p>\u003Cul>\u003Cli>ΔT\u003Csub>f\u003C/sub> = freezing-point depression (°C)\u003C/li>\u003Cli>K\u003Csub>f\u003C/sub> = cryoscopic constant of the solvent (for water, 1.86 °C/m)\u003C/li>\u003Cli>m = molality; i = van’t Hoff factor\u003C/li>\u003C/ul>\u003Cp>The new freezing point = normal freezing point − ΔT\u003Csub>f\u003C/sub>. This principle explains why salt (NaCl) is spread on icy roads — the dissolved ions depress the freezing point of water well below 0 °C. A 0.50 m aqueous glucose solution gives ΔT\u003Csub>f\u003C/sub> = (1)(1.86)(0.50) = 0.93 °C, so it freezes at −0.93 °C. Note that K\u003Csub>f\u003C/sub> for water is roughly four times larger than K\u003Csub>b\u003C/sub>, making freezing-point depression easier to measure experimentally.\u003C/p>",[33],[60],{"label":61,"body":62},"Why this matters: ice cream churns in salted ice","\u003Cp>The old-fashioned ice-cream maker is a freezing-point-depression machine: rock salt scattered on the ice jacket dissolves into the meltwater and drags the brine temperature well below 0 °C, cold enough to freeze the cream mixture (which, full of sugar and milk solids, has a depressed freezing point of its own). The same chemistry sets road-salting’s limits: a saturated NaCl brine bottoms out near −21 °C, so below roughly that temperature highway crews switch to calcium chloride or sand, because no amount of extra NaCl can push the freezing point lower than the saturation limit allows.\u003C/p>",{"heading":64,"content":65,"relatedObjectives":66,"deepDive":68},"Osmotic Pressure","\u003Cp>\u003Cstrong>Osmosis\u003C/strong> is the net flow of solvent through a semipermeable membrane from a region of lower solute concentration to higher solute concentration. The minimum pressure needed to halt this flow is the \u003Cstrong>osmotic pressure (Π)\u003C/strong>.\u003C/p>\u003Cp>The equation: \u003Cstrong>Π = iMRT\u003C/strong>\u003C/p>\u003Cul>\u003Cli>Π = osmotic pressure (atm)\u003C/li>\u003Cli>i = van’t Hoff factor\u003C/li>\u003Cli>M = molarity of the solution (mol/L)\u003C/li>\u003Cli>R = 0.08206 L·atm/(mol·K)\u003C/li>\u003Cli>T = temperature in kelvins\u003C/li>\u003C/ul>\u003Cp>Osmotic pressure is extremely sensitive to solute concentration, making it the preferred colligative property for determining molar masses of large molecules such as proteins, where ΔT\u003Csub>b\u003C/sub> or ΔT\u003Csub>f\u003C/sub> would be immeasurably small. In biology, solutions with equal osmotic pressure are called \u003Cstrong>isotonic\u003C/strong>; a solution with higher Π is \u003Cstrong>hypertonic\u003C/strong>; lower is \u003Cstrong>hypotonic\u003C/strong>. Intravenous (IV) fluids must be isotonic with blood to prevent cell damage.\u003C/p>",[67],147,[69],{"label":70,"body":71},"Why this matters: IV bags are osmotic engineering","\u003Cp>Hospital saline is 0.9% NaCl for an exact reason: that concentration is isotonic with your blood, producing the same osmotic pressure as the fluid inside red blood cells. Drip pure water into a vein instead and osmosis drives water into the cells until they swell and burst; an over-concentrated solution shrivels them by pulling water out. Reverse-osmosis desalination plants run the section’s equation from the other side: they apply pressure greater than seawater’s osmotic pressure (around 27 atm) to force water backwards through the membrane, leaving the salt behind. Π = iMRT prices out both the IV bag and the drinking water.\u003C/p>",{"heading":73,"content":74,"relatedObjectives":75,"deepDive":77},"The van't Hoff Factor","\u003Cp>The \u003Cstrong>van’t Hoff factor (i)\u003C/strong> accounts for the dissociation of electrolytes into multiple particles, which amplifies every colligative effect:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Nonelectrolytes\u003C/strong> (e.g., glucose, sucrose): i = 1 — molecules remain intact.\u003C/li>\u003Cli>\u003Cstrong>Strong electrolytes\u003C/strong>: i equals the total number of ions produced. NaCl → Na\u003Csup>+\u003C/sup> + Cl\u003Csup>−\u003C/sup>, so i = 2. CaCl\u003Csub>2\u003C/sub> → Ca\u003Csup>2+\u003C/sup> + 2 Cl\u003Csup>−\u003C/sup>, so i = 3.\u003C/li>\u003Cli>\u003Cstrong>Weak electrolytes\u003C/strong>: 1 < i < theoretical maximum, because dissociation is incomplete.\u003C/li>\u003C/ul>\u003Cp>In practice, measured i values for strong electrolytes are slightly \u003Cem>less\u003C/em> than theoretical due to \u003Cstrong>ion pairing\u003C/strong> — oppositely charged ions can temporarily associate in solution, reducing the effective particle count. This deviation grows at higher concentrations. For example, 0.10 m NaCl ideally gives ΔT\u003Csub>f\u003C/sub> = (2)(1.86)(0.10) = 0.37 °C, but the measured value is typically about 0.35 °C.\u003C/p>",[76],149,[78],{"label":79,"body":80},"Common mistake: counting atoms instead of ions","\u003Cp>The van’t Hoff factor counts the pieces a formula unit breaks into, not the atoms it contains. MgSO\u003Csub>4\u003C/sub> gives i = 2 (one Mg\u003Csup>2+\u003C/sup>, one intact SO\u003Csub>4\u003C/sub>\u003Csup>2−\u003C/sup>), not 6: the sulfate ion travels as a single particle, exactly as it has since the nomenclature topic. CaCl\u003Csub>2\u003C/sub> gives 3; glucose, which never dissociates, stays at 1 no matter how many atoms it holds.\u003C/p>\u003Cp>The two-step check: is the solute an electrolyte at all (molecular compounds mostly are not)? If so, how many \u003Cem>ions\u003C/em> per formula unit, keeping every polyatomic ion whole? Answer those two and i falls out correctly every time.\u003C/p>",{"heading":82,"content":83,"relatedObjectives":84,"deepDive":85},"Determining Molar Mass from Colligative Data","\u003Cp>Because colligative-property equations link a measurable physical change to the amount of dissolved solute, they can be rearranged to find an unknown \u003Cstrong>molar mass\u003C/strong>. The general strategy is:\u003C/p>\u003Col>\u003Cli>Measure the colligative effect (ΔT\u003Csub>b\u003C/sub>, ΔT\u003Csub>f\u003C/sub>, or Π).\u003C/li>\u003Cli>Use the appropriate equation to calculate molality (or molarity for Π).\u003C/li>\u003Cli>From molality and the known mass of solute and solvent, calculate moles of solute.\u003C/li>\u003Cli>Divide the mass of solute (in grams) by the moles to get the molar mass.\u003C/li>\u003C/ol>\u003Cp>Freezing-point depression is commonly used for small molecules because K\u003Csub>f\u003C/sub> is large enough to give measurable temperature changes. For macromolecules such as proteins or polymers, osmotic pressure is preferred — even very dilute solutions generate measurable Π values, allowing accurate molar-mass determinations where ΔT methods would fail.\u003C/p>\u003Cp>\u003Cstrong>From molar mass to molecular formula.\u003C/strong> Once colligative measurement gives the molar mass, a separate step combines it with percent composition to identify the molecular formula. The workflow: (1) percent composition → mole ratio of each element per 100 g → empirical formula and its empirical molar mass; (2) molecular molar mass ÷ empirical molar mass = n (a small whole number); (3) multiply empirical subscripts by n. Worked example: an organic compound with composition 93.46% C / 6.54% H gives the empirical formula C\u003Csub>6\u003C/sub>H\u003Csub>5\u003C/sub> (empirical MW 77.10 g/mol). If the colligative-property data yields a molar mass of 154.2 g/mol, then n = 154.2 / 77.10 = 2.00, so the molecular formula is (C\u003Csub>6\u003C/sub>H\u003Csub>5\u003C/sub>)\u003Csub>2\u003C/sub> = C\u003Csub>12\u003C/sub>H\u003Csub>10\u003C/sub> (biphenyl). For pure elements like sulfur, the molar mass is divided by the atomic mass to recover the allotrope subscript n (e.g., MW ≈ 256 g/mol ÷ 32.07 g/mol = 8, identifying S\u003Csub>8\u003C/sub>). The empirical-formula-to-molecular-formula derivation itself is the subject of \u003Cstrong>\u003Ca href=\"/learn/the-mole-and-chemical-formulas\">T8: The Mole and Chemical Formulas\u003C/a>\u003C/strong> — this topic uses it as the closing step in colligative-property molar-mass problems that ask for the molecular formula.\u003C/p>",[32,33,67],[86],{"label":87,"body":88},"Why this matters: weighing molecules with a thermometer","\u003Cp>Before mass spectrometers, this section’s rearrangement was a primary way to weigh a molecule: dissolve a known mass of the mystery compound, measure how far the freezing point drops, and the molality (hence moles, hence molar mass) falls out. Generations of newly discovered compounds got their molar masses from a thermometer and a balance.\u003C/p>\u003Cp>The method is not just history. Osmotic pressure, the most sensitive colligative effect, is still used to estimate molar masses of polymers and proteins: molecules so heavy that a solution’s freezing point barely moves, while its osmotic pressure remains comfortably measurable.\u003C/p>",{"heading":90,"content":91,"relatedObjectives":92},"Colligative Property Problem Workflow and Common Mistakes","\u003Cp>A reliable workflow for colligative property calculations:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Determine the solute type\u003C/strong>: molecular (i = 1) or electrolyte (i = number of ions produced per formula unit).\u003C/li>\u003Cli>\u003Cstrong>Calculate molality\u003C/strong> (not molarity) using moles of solute per kilogram of solvent.\u003C/li>\u003Cli>\u003Cstrong>Select the correct equation\u003C/strong>: ΔT\u003Csub>b\u003C/sub> = iK\u003Csub>b\u003C/sub>m for boiling-point elevation, ΔT\u003Csub>f\u003C/sub> = iK\u003Csub>f\u003C/sub>m for freezing-point depression, Π = iMRT for osmotic pressure.\u003C/li>\u003Cli>\u003Cstrong>Apply the van ’t Hoff factor\u003C/strong>. For strong electrolytes, i equals the number of ions per formula unit (NaCl → i = 2, CaCl\u003Csub>2\u003C/sub> → i = 3).\u003C/li>\u003Cli>\u003Cstrong>Check the direction\u003C/strong>: boiling points go up, freezing points go down. If your ΔT has the wrong sign relative to the pure solvent, recheck.\u003C/li>\u003C/ol>\u003Cp>Common mistakes: using molarity instead of molality for boiling-point and freezing-point equations, forgetting the van ’t Hoff factor for ionic solutes (this can double or triple the expected effect), treating a weak electrolyte as though it fully dissociates, and confusing the new boiling/freezing point with the ΔT value itself.\u003C/p>",[32,33,67,42,76],{"title":94,"problem":95,"steps":96,"answer":102},"Freezing Point Depression with an Electrolyte","Calculate the freezing point of a solution made by dissolving 10.0 g of CaCl₂ (molar mass 110.98 g/mol) in 250.0 g of water. Kᶠ for water = 1.86 °C/m.",[97,98,99,100,101],"Find moles of CaCl₂: 10.0 g ÷ 110.98 g/mol = 0.09011 mol.","Calculate molality: m = 0.09011 mol ÷ 0.2500 kg = 0.3604 m.","Determine the van’t Hoff factor. CaCl₂ → Ca²⁺ + 2 Cl⁻ produces 3 ions, so i = 3.","Calculate ΔTᶠ = i · Kᶠ · m = (3)(1.86)(0.3604) = 2.01 °C.","New freezing point = 0.00 °C − 2.01 °C = −2.01 °C.","The solution freezes at −2.01 °C. The three ions produced by CaCl₂ dissociation triple the colligative effect compared to a nonelectrolyte of the same molality.",[104],"OpenStax Chemistry 2e, Ch 11.4: Colligative Properties (CC BY 4.0)",[106,108,110,112,115,117,119,121,123],{"question":107},"What is a colligative property and why does it depend only on solute particle count?",{"question":109},"What is Raoult’s law and what does it predict?",{"question":111},"What is boiling-point elevation and what equation describes it?",{"question":113,"hint":114},"What is freezing-point depression and what equation describes it?","The equation is very similar to the one for boiling-point elevation.",{"question":116},"What is osmotic pressure?",{"question":118},"What is the van ’t Hoff factor (i) and why does it matter for electrolytes?",{"question":120},"How can colligative property data be used to determine the molar mass of an unknown solute?",{"question":122},"Why does an ionic solute like NaCl have a greater effect on colligative properties than a molecular solute at the same molality?",{"question":124},"What is the difference between an ideal and a non-ideal solution?",[126,127],"solutions-and-concentration","liquids-solids-intermolecular-forces",[129,130,131,132],"Na","Cl","H","O",[134],"colligative-constants",[136,139,142,145],{"label":48,"equation":137,"note":138},"ΔT_b = iK_b m","K_b for water = 0.512 °C/m; new bp = normal bp + ΔT_b",{"label":56,"equation":140,"note":141},"ΔT_f = iK_f m","K_f for water = 1.86 °C/m; new fp = normal fp − ΔT_f",{"label":64,"equation":143,"note":144},"Π = iMRT","M is molarity; R = 0.08206 L·atm/(mol·K); T in kelvin",{"label":146,"equation":147,"note":148},"Raoult's Law","P_solution = X_solvent × P°_solvent","X_solvent is the mole fraction of the solvent",{"title":150,"steps":151},"How to Calculate the Freezing Point of a Solution",[152,153,154,155,156],"Determine whether the solute is a nonelectrolyte (i = 1) or an electrolyte, and find its van't Hoff factor (i = number of ions produced per formula unit).","Calculate the molality of the solution: m = moles of solute / kilograms of solvent.","Apply the freezing-point depression equation: ΔT_f = iK_f m.","Subtract ΔT_f from the normal freezing point of the pure solvent to get the new freezing point.","Verify the result: the solution should freeze at a lower temperature than the pure solvent.",{"question":158,"answer":159,"type":160},"Two aqueous solutions have the same molality: one contains glucose (C6H12O6) and the other contains sodium chloride (NaCl). Which solution will have the lower freezing point, and by approximately what factor will the effects differ?","The NaCl solution will have the lower freezing point. NaCl is a strong electrolyte that dissociates into two ions (Na+ and Cl-), giving a van't Hoff factor of approximately 2. Glucose is a nonelectrolyte with i = 1. Since freezing-point depression depends on i × Kf × m, the NaCl solution produces roughly twice the freezing-point depression of the glucose solution at the same molality.","conceptual",[162,166],{"id":163,"problem":164,"type":165},"pt-19-1","Calculate the boiling point of a solution made by dissolving 15.0 g of NaCl (molar mass 58.44 g/mol) in 200.0 g of water. Kb for water = 0.512 °C/m. Assume i = 2.","calculation",{"id":167,"problem":168,"type":165},"pt-19-2","A protein is dissolved in water and the osmotic pressure is measured at 25 °C. If 5.00 g of protein in 1.00 L of solution gives an osmotic pressure of 1.54 × 10^-3 atm, estimate the molar mass of the protein.",{"workedExampleCount":170,"hasWorksheets":171},10,true,[173,174,175,176,177],"colligative properties","boiling point elevation","freezing point depression","osmotic pressure","van't Hoff factor","2026-07-16",5,{"id":181,"slug":182,"lesson":181,"title":183,"shortTitle":184,"description":185,"category":8,"objectiveCount":186,"problemCount":187},18,"intermolecular-forces-and-phase-changes","Intermolecular Forces and Phase Changes","IMFs & Phases","Identify and rank intermolecular forces, interpret phase diagrams and heating/cooling curves, classify crystalline solids, calculate the energy required for temperature and phase changes, and compute density and atomic radius from cubic unit-cell parameters.",7,55,{"id":189,"slug":190,"lesson":189,"title":191,"shortTitle":192,"description":193,"category":194,"objectiveCount":186,"problemCount":195},20,"chemical-kinetics","Chemical Kinetics","Kinetics","Study reaction rates, rate laws, reaction order, activation energy, and reaction mechanisms.","thermodynamics-kinetics",51,1785108608153]