Chem 110 Recitation – Week 3 – Solutions and Colligative Props.
Solubility – You may be asked whether your solute will be soluble in your solvent. Just figure out the intermolecular forces present in your solute and solvent, and then remember the phase LIKE DISSOLVES LIKE.
Solubility Samples:
1. Predict which solvent will dissolve more of the given solute:
(a) sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH)
Methanol
(b) Ethylene glycol (HOCH2CH2OH) in hexane or water
Water
(c) Diethyl ether (CH3CH2OCH2CH3) in water or ethanol (CH3CH2OH)
Ethanol
Dilutions–
M1V1 = M2V2
Dilution Samples:
1. What volume of 0.0380 M KMnO4 stock solution should we use to prepare 250 mL of 1.50x10-3 M KMnO4?
M1V1 = M2V2
(0.0380 M)(V) = (250 mL)( 1.50x10-3)
V = 9.87 mL
2. Calculate the volume of 0.0155 M HCl that we should use to prepare 100mL 5.23x10-4 M HCl.
M1V1 = M2V2
(0.0155 M)(V) = (100 mL)( 5.23x10-4)
V = 3.37 mL
Concentrations - You can qualitatively describe solutions in terms of dilute or concentrated. More importantly, there are a number of ways to quantitatively describe solutions. You will need to know these backwards and forwards, and will need to be able to get from one to another.
Molarity, M,
Molality, m,
Mass %,
Mole Fraction, X,
Concentration Samples:
1. Calculate the mass of glucose needed to prepare 150 mL of 0.442 M C6H12O6.
2. A sample of 0.892 g of potassium chloride is dissolved in 54.6 g water. What is the percent by mass of KCl in the solution?
3. The density of 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g/mL. What is the molality of the solution? The molar mass of methanol is 32.04 g/mol.
Assume 100 mL:
100 mL(0.976 g/mL) = 97.6 g(1kg/1000g) = .0976 kg
4. Calculate the molality of a 35.4 % (by mass) aqueous solution of phosphoric acid (H3PO4). The molar mass of phosphoric acid is 98.00 g/mol.
100g - 35.4g = 64.6 g solvent = 0.646 kg
5. A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What is the mole fraction of the solute?
142 g(1 mol/60.0952 g) = 2.363 mol solute
58.0 g(1 mol/18.01 g) = 3.22 mol solvent
6. Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution as a hair bleach. An aqueous solution of H2O2 is 30.0 % by mass and has a density of 1.11 g/mL. Calculate the molality, mole fraction and molarity.
Pieces of Information:
Mass solvent: 100g soln – 30.0 g solute = 70.0 g solvent = 0.070 kg
Moles solute: 30.0 g(1 mol/34.02 g) = 0.882 mol solute
Moles solvent:70.0 g(1 mol/18.01 g) = 3.88 mol solvent
Volume solution:100 g soln(1 mL/1.11 g) = 90.1 mL = 0.0901 L
Molality:
Mole Frac.:
Molarity:
Colligative Properties:
Vapor Pressure Reduction (VPR)
Raoult's Law:
vpsolution = (Xsolvent)(vppuresolvent)
vp = vapor pressure
vppuresolvent = a constant
Examples:
1. Calculate the vapor pressure of a 2.50% by mass aqueous ethylene glycol solution (HOC2H4OH) at 0oC. The vapor pressure is 4.58 torr at this temperature.
100 g solution – 2.50 g solute = 97.5 g solvent
97.5 g H2O = 5.42 mol H2O2.50 g E.G. = 0.0403 mol E.G.
18 g/mol62.07 g/mol
Mole Fraction = 5.42 mol/(5.42 mol + 0.0403 mol) = 0.993
P = (0.993)(4.58 torr) = 4.55 torr
2. Determine the change in the vapor pressure of water when 5.95 g of urea, CO(NH2)2, is dissolved in 100g of water at 10oC. (the vapor pressure of water at 10oC is 9.21 torr)
100 g water = 5.55 mol H2O5.95 g sucrose = 0..991 mol C12H22O11
18 g/mol 60.06 g/mol
Mole Fraction = 5.55 mol/(5.55mol + .991 mol) = 0.982
P = (0.982)(9.21 torr) = 9.04 torr
However, this one is asking for the change in pressure. So we need to take the original pressure of the water (9.21 torr) and subtract our new pressure from it.
P = 9.21 torr – 9.04 torr = 0.17 torr
Boiling Point Elevation (BPE)
Tb = Kb*cm
Freezing Point Depression (FPD)
Tf = Kf*cm
Kf and Kb = constants
cm = concentration (in molality)
Examples:
1. Estimate the freezing point depression and the freezing point of 0.10 m sucrose solution.
Tf = Kf*m
Tf = (1.86 k*Kg/mol)(0.10 m) = 0.19 oC or 0.19 K.
This number is the actual change though, which explains why it is positive. Since the question asks for the freezing point, you should subtract this number from the freezing point of water.
0oc – 0.19oC = -0.19oC
2. You add 1.00 kg of ethylene glycol antifreeze (C2H6O2) to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the solution?
Molality:
Tb = (0.512 C/m)(3.62 m) = 1.85oC 100 + 1.85 = 101.85oC
Tf = (1.86 C/m)(3.62 m) = 6.73oC 0.0 – 6.73 = -6.73oC
Osmotic Pressure (OP)
Examples:
1. What concentration of glucose, C6H12O6, will have the same osmotic pressure as blood (7.7 atm) at 25oC?
This may seem likes it's worded funny, but really it's not. You are looking for the concentration of glucose, in M, that will give a pressure of 7.7 atm at 25oC. Once you realize this, you can plug the numbers into the equation.
7.7 atm = (M)(0.0821 (L*atm)/(K*mol))(298 K)
M = 0.31 mol/L
2. What is the osmotic pressure of a solution formed by dissolving 50.0 mg of aspirin, C9H8O4, in 0.250 L of water at 25oC?
Just like the last one, only you have the added step of finding the Molarity of the aspirin solution.
Now, plug into our equation.
Pressure = MRT = (.0011 M)(.00821 (L*atm)/(K*mol))(298 K)
Pressure = 0.00270 atm
Determining a Solute's Molar Mass from it's Colligative Properties
1. When 8.05 g of an unknown compound X was dissolved in 100 g benzene, the vapor pressure of the benzene decreased from 100.0 torr to 94.8 torr at 26oC. What is (a) the mole fraction and (b) the molar mass of X?
First, you can use Raoult's law to solve for the mole fraction of the solvent:
Psolvent = (Xsolvent)(Posolvent) 94.8 torr = (X)(100 torr) = 0.948 torr
Second, you can use the mole fraction you found to solve for the moles of the solute and solvent. If you break "n" down to g/MM and plug all the numbers in, you should some up with one missing molar mass.
2. A 1.14 g sample of a molecular substance dissolved in 100 g of camphor freezes at 176.9oC, while camphor's normal boiling point is179.8. What is the molar mass of the substance, if the freezing constant for camphor is 39.7 K*Kg/mol?
First, find T.
179.8-176.9oC = 2.9oC or 2.9K
Second, plug the information into the freezing point depression equation, substituting g/MM for mol.
3. A 0.40 g sample of a polypeptide dissolved in 1.0 L of an aqueous solution at 27oC has an osmotic pressure of 3.74 torr. What is the molar mass of the polypeptide?
First, realize that your constant is in terms of atm, and you are given torr. You'll need to convert to atm before proceeding.
3.74 torr(1atm/760 torr) = 0.00492 atm
Second, you can plug into the osmotic pressure equation to get the correct answer, once again using g/MM instead of mol.
Solving for MM gives 2.0x103 g/mol