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Chemistry of Solutions, Lecture 3 Notes

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Chemistry of Solutions, Lecture 3 Notes 202-NYB-05

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These notes go over calculating the vapour pressure of solutions (Raoult's law), boiling-point elevation,and freezing-point depression.
Chemistry of Solutions
Nadia Schoonhoven
Class Notes
Raoult's Law, vapor pressure, Boiling Point Elevation, freezing point
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This 12 page Class Notes was uploaded by CatLover44 on Monday September 12, 2016. The Class Notes belongs to 202-NYB-05 at Dawson Community College taught by Nadia Schoonhoven in Fall 2016. Since its upload, it has received 8 views. For similar materials see Chemistry of Solutions in Chemistry at Dawson Community College.


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Date Created: 09/12/16
Chemistry of Solutions Course Number: 202-NYB-05 Lecture no. 3 Date: Thursday, September 1, 2016 Professor: Nadia Schoonhoven Topics Covered: determining molar mass from Raoult's law, solution composition, deviations from Raoult's law, vapour pressure, boiling-point elevation, and freezing-point depression. Exercise: Calculate the expected vapour pressure above a solution containing 158.0 g of sucrose (table sugar, molar mass = 342.3 g) in 643.5 mL of water at 25°C. At this temperature, the density of water is 0.9971 g/mL and the vapour pressure is 23.76 torr. Given: 158.0 g of sucrose, molar mass = 342.3 g, 643.5 mL water, 25°C, density of water at 25°C is 0.9971 g/mL. Calculate the expected vapour pressure above the solution. Raoult's law equation: P solutionxsolvent° solvent We want to calculate the value of P solution We are missing the mole fraction of water (x solvent 1. Calculate the moles of sucrose in 158.0 g. 158.0 g × (1 mol. / 342.3 g) = 0.46158 mol. sucrose. 2. Calculate the moles of water in 634.5 mL. 0.9971 g/mL × 643.5 mL = 641.6339 g water 641.6993 g × 1 mol. / 18.015 g = 35.6203 mol. water. 3. Calculate the mole fraction of water. 0.46158 mol. sucrose + 35.6203 mol. water = 36.082 mol.. X = 35.6203 mol. water/ 36.082 mol. solution water = 0.98720 4. Calculate the pressure above the solution: 0 P solutionxsolvent solvent (0.9872) × (23.76 torr) = 23.46 torr. Note: keep the following plot in mind when you're asked to solve problems involving Raoult's law: 2 The Importance of Raoult's Law Ø Raoult's law can be used to determine molar mass of a compound. If you're given the change in vapour pressure of a solution, and you're told that a certain amount (in grams) of a certain substance is added to some amount of a solvent, solve for x solventin the equation for Raoult's law. From here you should be able to solve for the number of moles of the compound you're interested in, so all you need to do at this point is convert moles to grams using the given information. Ø You can use Raoult's law to characterize compounds since variations in vapour pressures depend on the total number of solute particles in a solution. Vapour pressure tells you about the solute that was dissolved in a solution. Take note of whether a compound exhibits molecular or ionic dissolution, because counting particles is essential in problems like these. For example, NaCl is an ionic compound, and it produces Na ions + - and Cl ions (2 particles), so two moles of ions are produces for 1 mole of NaCl. On the other hand, sugar doesn't dissociate in water, so one mole of sugar produces only one mole of solute (1 particle). Exercise: Predict the vapour pressure of a solution prepared by mixing 35.0 g solid Na SO (mola2 4 mass = 142 g) with 175 g of water at 25°C. The vapour pressure of pure water is 23.76 torr at 25°C. P = x P° solution solvent solvent 1. Calculate the moles of water in 175 g. 175 g × 1 mol./ 18.015 g = 9.7141 mol. water. 2. Calculate the moles of Na SO i2 35.4 g. 35.0 g × 1 mol./ 142 g = 0.24648 mol. Na SO . 2 4 2- Use the equation: Na SO2( )4 s2Na( )+ +aqO ( ) 4 aq Three moles of solute per mole of Na SO , s2 3(0424648)= 0.73944 mol. Na SO . 2 4 3. Calculate the mole fraction of water. For total moles in solution, 9.7141 mol. water + 0.73944 mol. Na SO = 20.4495 mol. 3 X water= 9.7141 mol./ 10.4495 mol. = 0.92962. 4. Calculate the vapour pressure using Raoult's law. P = x P 0 solution solvent solvent P solution =.92962) × (23.76 torr) = 22.078 torr, round to 3 sig figs for the final answer: 22.1 torr. Liquid-Liquid Solutions We’ve applied Raoult's law to solutions with non-volatile solutes, like solutes with no vapour pressure, but Raoult's law can also be applied to solutions made of solvents with different vapour pressures. When you're faced with a problem that has two volatile components, both components add to the pressure’s solution, or P solutionso use the following equation when you're solving problems involving 2 volatile components: P solution(xAP° A + (xBP° B Where P solution still the observed pressure of the solution,Ax is the mole fraction of the first component, and P° isAthe partial pressure of A. The same logic extends to the second part of the equation, except you'd calculate the mole fraction of B and the partial pressure of B. A two volatile-component solution is only ideal if the components react as strongly as they do with each other as they would on their own. Consider the intermolecular forces and polarity of the components. For example, hexane (C H ) a6d14ctane (C H ) are8b18h non-polar, so they would share London dispersion interactions. The molecules of these compounds don't prefer to interact with themselves or each other, so the intermolecular interactions between octane and hexane are as strong as they would be if they were bonding with themselves. This constitutes an ideal two volatile-component solution (no deviation from Raoult’s law). It's very helpful to visualize the plots of reactions when you're dealing with Raoult's law because the graphs can tell you important information about the vapour pressures of the individual components of a solution and the solution itself. From this information, you can draw conclusions about boiling points and the solution’s composition, even if the problem you're faced with doesn't provide you with a lot of information. Keep the graphs in this lecture in mind when solving these problems. 4 Exercise: the vapour pressures or pure benzene and toluene at 25 degrees Celsius are 95.1 torr and 28.4 torr respectively. A solution is prepared in which the mole fractions of benzene and toluene are both 0.500. What are the partial pressures of the benzene and toluene above the solution? And what's the total vapour pressure of the solution? Benzene and toluene are both volatile. 0 0 Use the equation: P solution(xAP A + (x B )Bto calculate their partial pressures, and the pressure of the solution. The fact that there are 0.500 moles of benzene and toluene is very important. Any time you're doing a problem like this one, and you find the moles of one of the volatile components, just subtract the value you have from 1 to get the number of moles for the other component of the solution. (x P ) = (0.500) × (95.1 torr) = 47.6 torr A A (x P ) = (0.500) × (28.4 torr) = 14.2 torr B B 0 0 Psolution (xAP A + (x B )B= 47.6 torr + 14.2 torr= 61.8 torr How to know when to use Raoult's law? Ø Think about what's being asked of you in the problem; o What kind of information is given? Vapour pressure? o What are they asking for? Partial pressure? Total pressure of he solution? o These hints are meant to make it more simpler to distinguish between scenarios where you should use the equation for Raoult's law, or its variation. Positive and Negative Deviations from Raoult's Law • A positive deviation from Raoult's law is characterized by a solution pressure that yields a pressure that's higher than anticipated. In these cases, the mixing of these components isn't favoured, and they bond more effectively with themselves than with each other. • Negative deviations from Raoult's law are characterized by a solution pressure that is lower than expected. Mixing is highly favourable in these cases, and the components of these solutions interact more strongly with each other than with themselves. 5 • Negative and positive deviations are demonstrated by certain plots: • Always look at the structure and intermolecular interactions between a solution’s components. • In non-ideal solutions, molecules are being kicked out of the solution, which partly explains why un-mixing is favourable in these cases. • Solutions composed of components of similar strength are ideal. Two non-polar components can either produce a negative deviation from Raoult's law, or an ideal solution. • Below is a table that summarizes the deviations from Raoult's law. You can use it to predict whether or not a solution is ideal. 6 • Note: In problems where you are asked if a solution is ideal, and you're given a value for the solution's vapour pressure, calculate the vapour pressure of that solution, and compare it to the vapour pressure the problem provides you with. If the value you calculated is less than the one you were given, the lower vapour pressure is ideal. This problem then shows a positive deviation from Raoult's law Problem: mixing some solutes and solvents can lead to non-ideal solutions where Raoult's law does not predict the solution vapour pressure accurately. Which of the following scenarios best depicts this type of situation? 1. Two polar liquids when mixed have a good possibility of producing negative deviations (actual vapour pressure is less than expected) from Raoult’s Law. (This statement is true. If you examine the two graphs that are just above the table explaining deviations from Raoult's law, you’ll see that when two polar liquids are combined, mixing is highly favoured and a negative deviation from Raoult's law occurs. 2. When two polar liquids are mixed, they have a good possibility of producing positive deviations (actual vapour pressure greater than expected) from Raoult’s Law. (False because a mixture of two polar liquids yields a negative deviation from Raoult's law; the components of this kind of mixture react more strongly with each other than they do with themselves, so mixing is highly favoured). 3. When a non-polar liquid and polar liquid are mixed they have a good possibility of producing a negative deviation from Raoult’s law. (False because a mixture of a polar and non-polar liquid yields a positive deviation from Raoult's law; the components of this type of solution react stronger with themselves than they do with each other. Un-mixing is favoured). 4. None of these seem logical. (False because we know, from analyzing the plots for nonideal mixtures of two volatile liquids that the first statement is true). Exercise: a solution contains 2.43 g acetone (C H O3 m6lar mass = 58.08 g) and 3.95 g of carbon disulfide (CS , 2olar mass = 76.15 g). At 35°C, this solution has a total vapour pressure of 645 torr. Is this an ideal solution? The vapour pressure of pure acetone and pure carbon disulfide at 35°C are 332 and 515 torr, respectively. Both components of the solution are molecular and volatile. Acetone is polar, carbon disulfide is non-polar. 1. Calculate the moles of C H 3 a6d CS . 2 7 2.43 g × 1 mol. / 58.08 g = 0.04184 mol. C H 3. 6 3.95 g × 1 mol. / 76.15 g = 0.05187 mol. CS . 2 2. Calculate the total number of moles in the solution. ntotal 0.04184 mol. + 0.05187 mol. = 0.0937 mol. 3. Calculate the mole fractions of acetone and carbon disulfide. xA= 0.04183 mol./ 0.0937 mol.= 0.4465 xB= 0.05187 mol./ 0.0937 mol. = 0.5535 4. P = (0.4465) × (332 torr) + (0.5535) × (515 torr) = 433 torr (3 sig figs) total 433 torr < 645 torr, which means this solution is ideal, and the vapour pressure of 645 torr exhibits a positive deviation from Raoult's law. The components have opposite types of intermolecular interactions (CS i2 polar, and C H 3 i6 non-polar), so un-mixing is favoured. Exercise: If a litre of wine contains 120 mL ethanol & 880 mL water, what is the wine’s vapour pressure at 20°C ? We want to find the pressure of the solution (wine and water) at 20°C. Use Raoult's law, Psoln = (X P° ) + (X P° ). A A B B P solution Pethanol+ Pwater= X ethanol P° ethanol X water+ P°water We have 12% alcohol by volume, so: 8 12/ 100 = 0.12 x 1 L = 0.12 L of ethanol in the bottle of wine. What percentage of the wine is water? 1 L – 0.12 L = 0.88 L of water, so 88% of the wine is water. Calculate the moles of ethanol in the wine: The density of ethanol is 0.785 g/ mL, and the density of water is 1.000 g/ mL. For the moles of ethanol: 120 mL × 0.785 g/ mL × 1 mol. / 46.07 g = 2.048 mol. ethanol. For the moles of water: 880 mL × 1.000 g/ mL × 1 mol. / 18.05 g = 48.83 mol. water. Mole fraction of ethanol: XA= 2.048 mol. / 50.878 mol. = 0.04025. Mole fraction of water: XB= 48.83 mol. / 50.878 mol. = 0.9597. For the pressure of the solution: Psolution (0.04025)(48.39) + (0.9597)(17.535) = 18.597 mm Hg = 19 mm Hg (2 sig figs). Boiling-Point Elevation and Freezing -Point Depression • Colligative Properties: properties that only depend on the number of solute particles in a solution. o Boiling-point elevation,freezing-point depression, and osmotic pressure are colligative properties. o Remember that the melting point is the same as the freezing point! 9 • Boiling-Point elevation: a solution boils when its vapour pressure is the same as atmospheric pressure (1 atm); o Adding a solute to a solution lowers the vapour pressure of a solution, so more energy is required to boil that solution. This is boiling-point elevation. o The presence of a nonvolatile solute raises the boiling point of a solvent. o The degree of boiling-point elevation depends only on the total concentration of dissolved particles, and the solvent used. o This equation can be adapted to boiling-point elevation or freezing-point depression calculations: § ΔT = Km , where ΔT may represent boiling-point elevation (ΔT ) or solute b freezing-point depression (ΔT), fnd K can represent either the boiling-point constant (K b or the freezing-point constant (K). f § Remember that for ionic compounds, every ion acts as a solute particle. Exercise: a solution is prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution has a boiling point of 100.34°C. Calculate the molar mass of glucose, a molecular solid present as individual molecules in solution. K ofbwater is 0.51°C kg/mol. ΔT b K mb solute We want the molar mass of glucose. We know that the boiling point of water is 100°C, so: ΔT b 100.34°C – 100.00°C = 0.34°C. We are given the K ofbwater: 0.51°C kg/mol. ΔT b K mb soluteecomes m solute ΔT b K .b m solute (0.34°C) / (0.51°C kg/mol) = 0.66667 m glucose. To calculate the molar mass of glucose from its molality: 0.66667 mol/kg × 0.15 kg water = 0.1000005 mol. glucose. Molar mass of glucose = mass of glucose / moles of glucose. Molar mass = (18.00 g / 0.1000005 mol.) = 179.9991 g/mol = 180 g/mol (2 sig figs). Freezing-Point Depression 10 • The presence of a non-volatile solute decreases the freezing-point of the solvent. o The added solute prevents the solvent molecules from arranging themselves into a lattice structure, so freezing (and melting) occur at lower temperatures. Less energy is required to melt the new substance because the solute weakens the intermolecular forces between the solvent particles. • Freezing rate = melting rate at lower temperatures than in the pure solvent. Exercise: what mass of ethylene glycol (C H O , mola2 m6ss2= 62.1 g/mol), the main component of antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at -23.3°C? Assume the density of water is 1g/mL, and K of water is 1.86 fC kg/ mol. Use the equation ΔT = K m f f solute We are given ΔT = 1.f6 °C kg/ mol, molar mass = 62.1 g/mol, the freezing-point of the car’s radiator, -23.3°C, the density of water, 1 g/mL, and the volume of water that the ethylene glycol must be added to, 10.0 L. ΔT f 0°C – (-23.3°C) = 23.3°C. m = (23.3°C / 1.86°C kg/mol) = 12.527 mol. C H O kg wate2.6 2/ 12.527 mol. C H O2kg6wat2/ × 10.0 kg = 125.27 mol. C H O 2 6 2. For the required mass of C H O : 2 6 2 3 125.27 mol. C H O2× 62.2 g/mol. C H O = 777923 6 C2H O = 7.80 ×10 g2C 6 O2(3 sig figs). 2 6 2 Exercise: consider a bottle of red wine with approximately 12% alcohol (C H OH) by volume. 2 5 Calculate the melting point of the wine. Will it freeze? If the melting point is less than -15°C in absolute value, it won't be frozen. If it's greater than -15°C in absolute value, it will freeze. Since this is a continuation of a problem that was solved earlier, we can use the data we calculated again. 1 L – 0.12 L = 0.88 L of water, which equals to 0.88 kg of water (solvent). The density of ethanol is 0.785 g/ mL, and the density of water is 1.000 g/ mL. 11 120 mL × 0.785 g/ mL × 1 mol. / 46.07 g = 2.048 mol. ethanol. m = 2.048 mol. ethanol / 0.88 kg water = 2.327 m ethanol ΔT f Km f ethanol 1.86°C kg/ mol × 2.327 mol/ kg = 4.328°C = 4.3°C (2 sig figs) The freezing point of the wine is: ΔT f = 0C – 4.3°C = -4.3°C Since -4.3°C < -15°C, the wine will freeze. 12


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