Sample quiz and test questions – Chapter 4
I. Terms and short answers
1. Solid – solid phase boundary is obtained experimentally using
thermal analysis
2. Number of degrees of freedom at the triple point of a pure substance equals
F = 1 – 3 + 2 = 0
3. Molarity of the solution is defined by equation
[J] = nJ /V
4. Raoult’s Law is represented by equation:
* pJ = xJpJ
5. Chemical potential of a component of an ideal gas solution is given by equation
o mJ = mJ + RT ln aJ
6. Mixture that boils without changing the composition is called
azeotropic mixture
7. Kb in the equation DTb = KbbB is called
ebulioscopic constant
II. Graphs
1. Using the T(xA) diagram sketch the lines representing the composition of liquid and vapor for a high-boiling azeotrope. Label areas, lines and show the boiling temperature of A and B.
see fig. 4.26 page 144 and the lecture notes
2. Illustrate with the m(T) diagram how the solute lowers the freezing point of the solvent. Label axes, lines and points with appropriate symbols.
See fig 4.20 page 138 and the lecture notes III. Problems
1. What is the molality of an aqueous methanol solution whose mole fraction x(CH3OH) = 0.2?
Assume 1 mole of solution which gives the following: 0.2 mole of CH3OH and 0.8 mole of H2O mass of H2O = 0.8 mole x 18 g/mole = 14.4 g = 0.0144 kg molality = number of moles of CH3OH/mass of H2O = 0.2mole/0.0144 kg molality = 13.89 molal
2. Density of a 2.0 molar solution of glucose (C6H12O6) in water is 1.24 g/mL. What is the molality of this solution?
Assume 1 L of solution. Mass of solution = density x volume = 1.24 g/mL x 1000 ml = 1240 g Mass of glucose = number of moles x molar mass = 2 x 180.16 = 360.32 g Mass of solvent = 1240 g – 360.32 g = 879.68 g = 0.88 kg Molality = 2 moles/0.88 kg = 2.27
3. Estimate the elevation of the boiling point and the osmotic pressure of the solution made by dissolving 5 g of MgCl2 in 300 mL of water.
Molar mass of MgCL2 = (24.3 + 2 x 35.05) g/mol = 94.4 g/mole 5 g/ 94.4 g/mole = 0.053 mole mass of H2O = d x V = 1 g/mL x 300 mL = 300 g = 0.3 kg molality of MgCl2 = 0.053/0.3 = 0.177 -1 Kb(H2O) = 0.51 (K kg mol )
DTb = Kb x 3 x m(MgCl2) = 0.51 x 3 x 0.177 = 0.27 K Assume that adding MgCl2 does not change the volume so the volume of the solution = 300 mL molarity of MgCl2 = 0.053/0.3 L = 0.177 p = 3 x molarity(MgCl2) x R x 298 K = 3 x 0.177 mol/L x 0.082 L atm/(K mol) x298K p = 12.98 atm = 1.32x106 Pa
4. 5 g of certain compound dissolved in 800 g of benzene lowered the freezing point of the solvent by 0.6 oC. Calculate the molar mass of the compound.
Molality of a compound = DT/Kf = 0.6/5.12 = 0.1172 Molality = n/m(kg) à 0.1172 = n/0.8 à n = 0.8 x 0.1172 = 0.094 mole Molar mass = 5 g/ 0.094 mole = 53.2 g/mole