Module 2 - Solutions

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Which of the following statements is true?

a) osmotic pressure is pressure from particles traveling through a membrane
b) osmotic pressure is the pressure required to force solvent particles to pass through a membrane
c) osmotic pressure is the pressure required to separate the solvent from the solute
d) osmotic pressure is the pressure required to stop osmosis
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Terms in this set (47)
Which of the following statements is true?

a) osmotic pressure is pressure from particles traveling through a membrane
b) osmotic pressure is the pressure required to force solvent particles to pass through a membrane
c) osmotic pressure is the pressure required to separate the solvent from the solute
d) osmotic pressure is the pressure required to stop osmosis
d) osmotic pressure is the pressure required to stop osmosis

Definition of osmotic pressure
Which of the following aqueous solutions would have the highest boiling point?

a) 1.0 m NaNO3
b) 1.0 m HF
c) 1.0 m C6H12O6
d) 1.0 m CaCl2
d) 1.0 m CaCl2

NaNO3 would dissociate into Na^+ and NO3^-, HF does dissociate, but CaCl2 dissociates into Ca^2+ and 2Cl, which has more ions inside and therefore a higher boiling point. C6H12O6 doesn't dissociate.

As a general rule, if the molecule begins with carbon it doesn't dissociate.
C6H14 is soluble in C5H12. In H2O, this solute would be

a) more soluble
b) less soluble
c) equally soluble
b) less soluble

C6H14 and C5H12 are both nonpolar. Like dissolves like.
What is the concentration in %w/v of a 0.425 M aqueous solution of methanol (MM = 32.04 g/mol)?

a) 136%
b) 13.6%
c) 1.36%
d) 0.136%
c) 1.36%

%w/v is defined as the number of grams per solute per 100 mL of solution, and can be calculated as (g solute)/(mL solution) x 100

Use molarity to calculate the grams of solute per mL of solution

0.425 mol methanol / 1L solution x 32.04 g / 1 mol x 1L / 1,000 ML = 0.01362 x 100 = 1.36%
Which of the following is an example of a colloid?

a) milk
b) coffee
c) soda
d) saltwater
a) milk

Homogenized milk contains large particles that don't settle out of solution.
Oxygen gas is most soluble in water under which set of conditions?

a) low temperature, low pressure
b) low temperature, high pressure
c) high temperature, high pressure
d) high temperature, low pressure
b) low temperature, high pressure

This principle is the same for any gas being dissolved in water.
The phrase "like dissolves like" refers to

a) two molecules having the same chemical formula
b) two molecules having the same molar mass
c) two molecules having the same intermolecular forces
d) two molecules having the same temperature
c) two molecules having the same IMFs

Polar substances dissolve into polar solvents; nonpolar substances dissolve into nonpolar solvents
A 83.5 g sample of a nonelectrolyte is dissolved in 325.3 g of water. The solution is determined to have a boiling point of 102.3 degrees C. What is the molar mass of the compound? (Kb for water is 0.510 degrees C/m)
Mm solute = Kb(g solute)i / (delta)Tb(kg solvent)

^note the above formula is exactly the same for freezing point, except use Kf.

The normal boiling point of water is 100 degrees C.

(delta)Tb = 2.3 degrees C

(0.510 degrees C/m)(83.5 g) / (2.3 degrees C)(0.3253 kg water) = 56.92 g
What is the molarity of an aqueous solution that contains 0.072 g C2H6O2 per gram of solution? The density of the solution is 1.04 g/mL.
(0.072 g C2H6O2 / 1 g sol'n) x (1.04 g sol'n /1 mL sol'n) x (1 mL / 0.001 L) x (1 mol / 62.07 g C2H6O2)

= 1.21 M
At a particular temperature, the solubility of O2 in water is 0.0500 M when the partial pressure is 0.120 atm. What will the solubility be when the partial pressure of O2 is 0.480 atm?
Henry's law

Cgas = KP

Use the given M and P to solve for the K value in this situation, and use that K value and the given atm to find Cgas

0.0500 atm = (0.120 atm)K
K = 0.416

Cgas = (0.416)(0.480 atm)

Cgas = 0.2 M
MgCl2 has a van't Hoff factor of 2.70. What would be the boiling point of an aqueous solution containing 5.10 mol of MgCl2 in 1.00 kg of water? (Kb for water is 0.51 degrees C/m)(delta)Tb = Kb(grams solute)i/Mm solute(kg solvent) Use the given mol of MgCl2 and the molar mass to find how many g of solute are in this particular solution 5.10 mol MgCl2 x (95.21 g MgCl2 / 1 mol MgCl2) = 485.571 g MgCl2 = (0.51)(485.571 g)(2.70) / (95.21 g)(1.00 kg) (delta)Tb = 7.0227 degrees C BP sol'n = BP solvent + (delta)Tb = 100 + 7.0227 BP sol'n = 107 degrees CThe driving factor for solution formation is a) entropy b) enthalpy c) temperature d) molar mass of the moleculesa) entropy Remember that entropy is a measure of the amount of disorder in a systemA liquid typically boils at 200 degrees C. If an insoluble salt is added to the liquid, its new boiling point will be: a) greater than 200 degrees C b) less than 200 degrees C c) 200 degrees C d) not enough information is provided to answer the questionc) 200 degrees C the boiling point would only be affected if the salt dissolved.How do solutions differ from compounds? From other mixtures?- a solution has the potential to vary in composition; a compound cannot - solutions are homogeneous at the molecular level; other mixtures are heterogeneous at the molecular levelWhen KNO3 is dissolved in water, the resulting solution is significantly colder than the water was originally. a) Is the dissolution of KNO3 an endothermic or exothermic process? b) What conclusions can you draw about the intermolecular attractions involved in the process? c) Is the resulting solution an ideal solution?a) Endothermic - the solution is consuming heat rather than releasing it b) The attraction between K^+ and NO3^- ions is stronger than the ions' attraction to H2O. The dissolution process therefore increases the energy of the molecular interactions, it consumes the thermal energy of the solution to make up for the difference. c) No - an ideal solution has no appreciable heat release or consumptionIndicate the most important types of intermolecular attractions in each of the following solutions: a) Potassium dichromate, K2Cr2O7, in H2O (l) b) NO (l) in CO (l) c) Cl2 (g) in Br(l) d) HCl (aq) in benzene, C6H6 (l) e) Methanol, CH3OH (l) in H2O (l)a) ion - dipole forces b) dipole - dipole forces (both are polar) c) dispersion forces d) dispersion forces e) hydrogen bondingHeat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes.- heat is released when the total IMFs between solute and solvent molecules are stronger than the total forces on the pure solute and in the pure solvent. Breaking weaker forces and forming stronger ones releases heat. - heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and the pure solvent. Breaking stronger forces and forming weaker IMFs absorbs heat.Explain why the ions Na^+ and Cl^- are strongly solvated in water, but not in hexane, a solvent composed of nonpolar molecules.Crystallized NaCl dissolves in water, and the ions are attracted to the water because H2O is highly polar. Hexane is nonpolar, so there's no charges for the ions to be attracted to. Remember - "like dissolves like"What is the expected electrical conductivity of the following solutions? a) NaOH (aq) b) HCl (aq) c) C6H12O6 (aq) d) NH3 (aq)a) high it would dissociate when dissolved into Na^+ and OH^- the solute is an ionic compound b) high the solute is a strong acid and would ionize completely when dissolved c) non-conductive solute is a covalent compound, neither acid nor base, so it is nonreactive towards water d) low solute is a weak base and will partially ionize when dissolvedWhat is the molality of phosphoric acid, H3PO4, in a solution of 14.5 g of H3PO4 in 125 g of water?remember - molality is the moles of solute divided by kg of solvent Steps: 1. Convert 14.5 g of H3PO4 from g to moles using the given mass and the molar mass. 2. Convert the mass of the water from g to kg 3. Divide the answer for 1 by the answer for 2 molar mass of H3PO4: 3(1.01) + 30.97 + 4(16.00) = 98.00 g/mol 14.5 g H3PO4 x (1 mol/98.00 g) = 0.1479... 125 g H2O = 0.125 kg 0.1479/0.125 = 1.184 mol/kgWhat is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO3 by mass)?Steps: 1. Determine the mass of HNO3 and water in 100 g of solution. 2. Convert the mass of HNO3 to moles 3. Convert the mass of water to kg 4. Divide 2 by 3 In 100 g of solution, 68.0 g are of HNO3 100-68 = 32.0 g of water molar mass of HNO3: 1.01 + 14.01 + 3(16.00) = 63.01 68.0 g HNO3 x (1 mol/63.01 g) = 1.08 mol 32.0 g H2O = 0.0320 kg H2O 1.08 mol/0.0320 kg = 33.8 mol/kgTyndall effectScattering of a light beam as it passes through a colloidColloid vs Solution vs SuspensionColloid: - medium sized particles - can be separated by semi-permeable membranes - cannot be filtered - less homogenous - uniform in terms of physical properties Solution: - transparent - does not separate - small particles, ions or molecules that cannot be filtered - different physical properties depending on how much the solute settles Suspension: - large particles that settle out - can be filtered - must be stirred to stay suspended (i.e muddy water, blood platelets)soluteA substance that is dissolved in a solution.solventA liquid substance capable of dissolving other substancesmiscibleliquids that dissolve each otherimmiscibleliquids that do not dissolve each otherexothermicheat is releasedendothermicheat is absorbedsaturatedall solute that can dissolve is dissolvedunsaturatedmore solute can be dissolvedsupersaturatedmore than usual is dissolvedsolvationsolvent particles bond to solute particleshydrationsolvation by waterSolution process**the solute must be attracted to the solvent** 3 attractions take place: - solute-solute - solvent-solvent - solute-solvent the first two are endothermic, the last is exothermic. the heat of solution is the addition of all three of these reactions, (note exothermic is negative), the sign of the sum determines whether the overall process was endo/exothermiccarbonation/Henry's lawdissolving CO2 (g) in water CO2 is nonpolar H2O is polar In order to force them to dissolve, you have to use pressure. The more pressure is applied, the more it dissolves. This is Henry's law: C=kP k = Henry's law constantConcentration unitsMass % = mass solute(100)/(mass solute + mass solvent) Molarity = moles solute/liter solution Molality = moles solute/kg solventSolutions used in hospitals usually contain 500 mL of saline solution, made of 4.5 g of NaCl in 490 g of water. What is the mass %, molarity, and molality?mass % = 4.5g (100) / (4.5 g + 490 g) = 0.91% molarity = 4.5 g x (1 mol/58.5 g) = 0.0769 mol / 0.49 L = 0.157 mol/L molality = 0.0769 mol/ 0.49 kg = 0.157 mol/kgHow many grams of NaOH are needed to make 150 g of 6.00% NaOH sol'n?In a 100 g solution, there is 6 g of NaOH 150 g soln x (6.00 g NaOH/ 100 g sol'n) = 9.00 g NaOHA student evaporates 55 mL of of 1.6 M KCl sol'n. How many grams of KCl remain after evaporation is complete?molarity is given, use it to convert from mL to mol then use the molar mass of KCl to convert from mol to g. 0..055 L x (1.6 mol/ 1 L) = 0.088 mol KCl 0.088 mol x (74.6 g / 1 mol) = 6.6 g KClWhat is the molality of a 12% C6H12O6 aqueous solution?100-12 = 88 g of water 0.088 kg of solvent 12 g C6H12O6 x (1 mol / 180.16 g) = 0.066607 mol/0.088 = 0.76 mol/kgcolligative propertiesProperties that depend on the number of solute particles (molecules, ions) in solution 1. Freezing point depression 2. Boiling point depression 3. Osmotic pressureDoes dissolving something into a solvent change its physical properties?Yes; solvent properties are affected by the solute. Freezing point tends to be more affected than boiling point.Does adding salt raise or lower the freezing point of water? Does adding antifreeze raise or lower the freezing point of water?Lower for both - this is why salt is laid on the roads during the winter, and why you should treat your car with antifreeze in the winter so the water in the radiator doesn't freezeFreezing Point DepressionDissolving a solute in a solvent lowers the freezing point of the solvent. (delta)Tf = Kf(grams of solute)i / Mm solute(kg solvent) FP sol'n = FP solvent - (delta)TfBoiling Point ElevationA dissolved solute raises the boiling point of the solvent (delta)Tb = Kb(grams of solute)i / Mm solute(kg solvent) BP sol'n = BP solvent + (delta)TbWhat is the osmotic pressure of a liter of solution at 25 degrees C that contains 5.85 g NaCl?osmotic pressure = MRTi Need to find the molarity, use the given amount of grams of NaCl 5.85 g NaCl x (1 mol/58.44 g NaCl) = 0.1 mol = 0.1 M (since the solution is already a liter) 25 degrees C = 298.15 K i = 2 because NaCl dissociates into 2 ions when dissolved in water osmotic pressure = MRTi = (0.1)(0.0821)(298.15 K)(2) = 4.895 atm