Honors Chemistry

Answers to Chapter 11 Study Questions

1.  sodium chlorate  = NaClO3 =  0.320 moles

2. = 1.94 moles = 20.7 moles

total moles = 1.94  + 20.7 = 22.6 moles; = 0.0857

3. =  30.0% KOH

30.0% KOH  =  30.0 g KOH + 70.0 g H2O;  molality = moles KOH/kg water =  7.64 moles KOH/kg water

4.  methanol = CH3OH

molarity  = ;    12.8 g CH3OH =  0.400 mol CH3OH

volume CH3OH = = 16.2 mL;  volume water = 144 mL

total volume of solution = 16.2 mL + 144 mL = 160 mL = 0.160 L

molarity = =  2.50 M

5.  The solubility of gases decreases as temperature increases.  Two everyday examples of this are:  1)  that soda becomes “flat” faster at room temperature than in the refrigerator since the solubility of CO2 is lower at 25°C than at 4°C, and 2) as water is heated, well before it boils, bubbles of air appear, since the solubility of air is decreasing during heating.

6.       Add a small crystal.  If it dissolves, the solution was unsaturated.  If it doesn't dissolve, the solution was saturated.  If more than the crystal comes out of solution, then the solution was supersaturated.

7.  DTf = 1.86 °C x moles solute particles/kg water =  3.72°C;    Tf  =  0 - DTf  =  0 - 3.72°C  =  -3.72°C

8.  DTf  = 1.86 °C x moles solute particles/kg water ;  calcium chloride = CaCl2 = 0.751 mol particles =  5.59°C;      Tf  =  -5.59°C

(CaCl2 is an electrolyte and it is important to remember that there are 3 moles of ions per mole of CaCl2.  The freezing point is three times lower than it would be for a nonelectrolyte.)

9.  molar mass  =  mass/moles;  mass = 80.0 g; find moles =  0.500 moles

molar mass =  80.0 g/0.500 moles  =  160. g/mole

10.  From Table 11.5, for benzene:  Kb = 2.53°C/m and Tb  =  80.1°C. =  6.32°C;  Tb  =  80.1°C + 6.32°C  =  86.4°C

11.  molar mass  =  mass/moles;  mass = 45.0 g; find moles; DTb  =  90.2°C - 80.1°C = 10.1°C =  0.300 moles

molar mass =  45.0 g/0.300 moles  =  150. g/mole