CHEM 1125Q 1st EditionExam # 2 Study Guide Lectures: 10-16Lecture 10 (February 25)Properties of SolutionsDescribe the 3 types of solutions.Unsaturated:- In an unsaturated solution, there is less solute than the solution is able to hold.Saturated:- In a saturated solution, the solution contains the maximum amount of solute that the solvent is able to hold at a certain temperature.Supersaturated:- In a supersaturated solution, the solution contains more solute than the solvent can hold, and is considered metastable.- Supersaturated solutions form when a saturated solution is slowly cooled so that the solubility is lower, but no precipitate forms.- The supersaturated solution is metastable because as long as it is not disturbed, it will not precipitate. If more solute is added, then precipitate will form.Concentration UnitsDistinguish between molarity, molality, and percent by mass.Molarity:- The Molarity of a solution is the moles of solute relative to the volume of solution.- M = (mols solute)/(L solution)Molality: - The Molality of a solution is the moles of solute relative to the mass of solvent.- m = (mols solute)/(kg solvent)Percent by Mass- The percent by mass of a solution is the mass of solute relative to the mass of solute and solvent.- % = (mass solute)/(mass solute+solvent) x 100Lecture 11 (March 2) Solubility and Henry’s LawUsing Henry’s Law, determine the molarity of a gas inside an unopened bottle of soda, with a pressure of 4.85atm and a K value of 3.05x10-2 M/atm.- Since we know that Henry’s Law is defined by C = KP, and we are given K and P, all we have to do is solve for C.3.05x10-2 M/atm x 4.85atm = 0.148Boiling Point Elevation and Freezing Point Depression18mol of a solute is added to 1kg of pure water. If KF for pure water is 1.86 °C/m, solve for the new freezing point of the solution.- For a Freezing Point Depression problem, we will be using the following equations: ΔTF = KFm and ΔTF = TF0 - TF. We know that m stands for molality, or mol/kg. We are given both the numbers required to solve for m as well as the KF. Therefore, we will first solve for ΔTF.18mol/1kg = 18m18m x 1.86 °C/m = 33.48°C = ΔTF- Next, we need to solve for TF . We know that TF0 for pure water is 0°C.0°C – 33.48°C = -33.48°C-33.48°C is the new freezing point.Lecture 12 (March 4)Osmotic PressureA 2.14M solution is heated to 315K. What is the osmotic pressure of the solution?- The equation we will use for this problem is: π = MRT. We are given T and M, and we know that R stands for the gas constant 0.08206 (L atm/mol K.) All we need to do is solve for the osmotic pressure.2.14M x 0.08206 (L atm/mol K) x 315K = 55.3atmElectrolyte SolutionsA 0.55M solution is heated to 300K and has an osmotic pressure of 4.05atm. Find the Van’t Hoff factor.- The equation we will use for this problem is π = iMRT. We are given all values needed to solve for i. Remember that R stands for the gas constant.4.05atm = i x 0.55M x 0.08206 (L atm/mol K) x 300Ki = 0.299Lecture 13 (March 9)EntropyFind the entropy for a system with 627 different arrangements.- For this problem we will use the equation: S = k ln(W). We are given W and we know that k stands for the Boltzmann constant, which is 1.38x10-23 J/K. All we have to do is solve for S.1.38x10-23 J/K x ln(627) = 8.89x10-23 J/K = SEntropy Change1.26mol of a gas expands from 1L to 5L. What is the entropy change for the system?- The equation we will use for this problem is ΔSsystem = nR ln(Vfinal/Vinitial). We are given both volumes as well as the number of moles, so all we need to do is solve for the entropy change. Remember that for this equation, the constant R represents the value 8.31 J/mol K.1.26mol x 8.31 J/mol K x ln(5L/1L) = 16.9 J/K = ΔSsystemLecture 14 (March 11)Qualitative Entropy PredictionsWhat are some processes that lead to an increase in entropy?- Melting, vaporization, and sublimation- Increase in temperature- Any reaction resulting in an increase in the number of gas moleculesCalculating Entropy of SurroundingsA system has an enthalpy change of -25kJ/mol at 298K. Find the entropy of the surroundings.- For this problem we will use the following equation: ΔSsurroundings = -ΔHsystem /T. Since we are giventhe temperature and enthalpy change, all we have to do is solve for entropy. Remember that enthalpy is in kJ while entropy is in J, so first we will have to convert.-25kJ/mol (1000J/1kJ) = -25,000J/mol-(-25,000J/mol) / 298K = 83.9J/mol K = ΔSsurroundingsLecture 15 (March 23)3 rd Law of ThermodynamicsDescribe the 3rd Law of Thermodynamics.- The 3rd Law of thermodynamics states that a perfect crystalline substance, which is a theoretical substance only, has an entropy of 0 at Absolute Zero, which is a theoretical minimum. As a substance’s temperature increases from Absolute Zero its entropy increases.Gibbs Free EnergyGiven an enthalpy change of -130kJ/mol and an entropy change of 980J/mol at 315K, find the Gibbs free energy change.- For this problem we will be using the equation: ΔG = ΔH – TΔS. Since we are given the valuesfor ΔH, T, and ΔS, all we have to solve for is ΔG. But first we have to make sure we have all the same units. Since ΔG is measured in kJ, we will convert the J of ΔS to kJ.980J/mol (1kJ/1000J) = 0.98 kJ/mol-130kJ/mol – (315K x 0.98kJ/mol) = -438.7kJ/molAccording to the value we got for ΔG above, is that process spontaneous or not spontaneous?- It is spontaneous because a process is spontaneous when ΔG < 0.Lecture 16 (March 25)No new material was covered in this