Practice Test Solids and Solutions (Part of Ch 11 and Ch 12)
1. Identify each as molecular, ionic, or atomic. If atomic, which type is it?
A. CaCl2 (s)
B. Ni (s)
C. CO2 (s)
d. I2 (s)
A. Ionic B. metallic/atomic C. molecular D. molecular
2. For the solids in question 1, which are conductive and why?
Just B because it is metallic and has delocalized electrons
A is not ionic because ions are in fixed positions
3. Which solid in question 1 would have the highest mp?
Most likely A because ionic/Coulomb strong IM
4. Which solid has the lower mp and why? H2O (s) or H2S (s); RbCl (s) or KCl (s); RbCl (s) or MgCl2 (s)
H2S because it cannot H bond
RbCl-Rb is larger than K so the Coulomb Forces are not as strong
RbCl: charge is less than in MgCl2 (Coulomb weaker)
5. Which of the following pairs of elements is most likely to create an interstitial alloy?
A. Ti and Cu
B. Al and Pb
C. Ag and Sn
D. Mg and Ca
Interstitial-when add a metal into empty spaces in metallic solid. Want to be mixing one
small with one larger (the larger would be original metallic solid): greatest size difference is
between Al and Pb (B)
6. Which of the following pairs of elements is most likely to create a substitutional alloy?
A. Ti and Cu
B. Al and Pb
C. Ag and Sn
D. Mg and Ca
Want two metals that are similar in size since one will be replacing the other. Maybe A
7. For the alloy pair you picked in question 5, how would the mp of the alloy compare to the mp of
the original metallic solid?
The alloy would be stronger/higher mp because packing density increases when add atoms to
the interstitial space
8. Explain the energy of the valence band compared to the energy of the conductance band for an
insulator? How does that change for a conductor? Insulator: they would be far apart (large
energy difference). As energy gap decreases, become a better conductor.
9. Is there any way to make something more conductive? Why or why not?
Anything that causes the band gap to decrease will make something more conductive. Doping is a
way to make semiconductors more conductive for example.
10. The phrase likes dissolves like refers to the fact that
A. Gas can only dissolve other gases.
B. Solvents can only dissolve solutes with similar intramolecular forces but different
intermolecular forces
C. Solvents can only dissolve solutes of similar molar mass
D. Solvents can dissolve solutes with similar intermolecular forces
11. An unsaturated solution is one that
A. Has no solute particles
B. Contains the maximum concentration of solute possible and is in equilibrium with
undissolved solute
C. Has a concentration lower than the solubility
D. Contains more dissolved solute than the solubility allows
12. Pressure has an appreciable effect on the solubility of ____ in liquids
A. Gases only
B. solids only
C. liquids only
D. all states of matter
E. solids and liquids only
13. SKIP this question for now (we need to do equilibrium before we can answer)
As temperature increases, the solubility of a solid in water
A. Increases if dissolution of the solid in water is exothermic
B. Increases if dissolution of the solid in water is endothermic
C. Increases regardless of the enthalpy
D. Decreases
E. Stays the same
14. Ammonia is soluble in water due to which solute-solvent force of attraction?
A. London dispersion-dipole
B. Ion-dipole
C. H- bonds
D. London dispersion forces only
15. What mass of aluminum nitrate is required to prepare 500.0 mL of a solution that is 0.100 M?
0.100 M * 0.500 L * MM Al(NO3)3 = mass of Al(NO3)3 (3 SFs)
16. A solution was prepared by dissolving 23.7 g of CaCl2 in 375 g water. The density of the resulting
solution was 1.05 g/mL. What is the [Cl-] in the solution?
23.7 g CaCl2 / MM = 0.214 mol CaCl2 yields twice as much Cl- = 0.428 mol ClTotal mass of solution is 375 g + 23.7 g = 398.7 g solution
Use density and mass of solution to find volume solution = 380. mL
Concentration of Cl- = 0.428 mol/0.380 L = 1.12 M
17. Calculate the M and mole fraction of methanol in a solution composed of 26.5 g methanol and
244 g water. Assume the density of the solution is 0.975 g/mL.
M = moles methanol/L solution
Convert 26.5 g methanol to mol (/MM) = 0.828 mol CH3OH
Total mass of solution = 270.5 g solution
Use density and mass to find volume of solution = 277 mL
M = 0.828 mol/0.277 L = 2.99 M
Mol fraction = mol methanol/total mol
Need mol of water to get total mol solution: 244 g water/18 = 13.56 mol water
Total mol = 13.56 + 0.828 = 14.39 mol total
Mole fraction methanol = 0.828/14.39 = 0.0575 (no units)