# Answers to Chapter 11 Study Questions

1.  sodium chlorate  = NaClO3

$$\mathrm{284\: g\: solution \times \dfrac{12.0\:g\:NaClO_3}{100\:g\:solution}\times\dfrac{1\:mol\:NaClO_3}{106.4\:g\:NaClO_3}=0.320\: moles}$$

2.  $$\mathrm{mole\: fraction =\dfrac{moles\:C_2H_6O_2}{total\:moles}}$$;  $$\mathrm{moles\: C_2H_6O_2 = 120.0\: g\times\dfrac{1\:mol\:C_2H_6O_2}{62.0\:g\:C_2H_6O_2}=1.94\: moles}$$

$$\mathrm{moles\: C_3H_6O = 1.20\: kg\times\dfrac{1000\:g}{1\:kg}\times\dfrac{1\:mol\:C_3H_6O}{58.0\:g\:C_3H_6O}=20.7\: moles}$$

total moles = 1.94  + 20.7 = 22.6 moles;  $$\mathrm{mole\: fraction =\dfrac{1.94\:mol\:C_2H_6O_2}{22.6\:moles}=0.0857}$$

3.  $$\mathrm{\dfrac{6.90\:mol\:KOH}{1\:L\:solution}\times\dfrac{1\:L}{1000\:mL}\times\dfrac{1\:mL}{1.29\:g}\times\dfrac{56.1\:g\:KOH}{1\:mol\:KOH}=\dfrac{0.300\:g\:KOH}{g\:solution}=30.0\%\: KOH}$$

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

$$\mathrm{\dfrac{30.0\:g\:KOH}{70.0\:g\:H_2O}\times\dfrac{1000\:g\:H_2O}{1\:kg\:H_2O}\times\dfrac{1\:mol\:KOH}{56.1\:g\:KOH}=\dfrac{7.64\: moles\: KOH}{kg\: water}}$$

4.  methanol = CH3OH

$$\mathrm{molarity=\dfrac{moles\:CH_3OH}{L\:solution}}$$;    $$\mathrm{12.8\: g\: CH_3OH\times\dfrac{1\:mol\:CH_3OH}{32.0\:g\:CH_3OH}=0.400\: mol\: CH_3OH}$$

$$\mathrm{volume\: CH_3OH = 12.8\: g\times\dfrac{1\:mL}{0.791\:g}=16.2\: mL}$$;  volume water = 144 mL

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

$$\mathrm{molarity =\dfrac{0.400\:mol\:CH_3OH}{0.160\:L\:solution}=2.50\:M}$$

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.  ΔTf = 1.86 °C x moles solute particles/kg water

$$\mathrm{\Delta T_f = 1.86\, ^\circ C \times\dfrac{0.11\:moles}{0.055\:kg\:H_2O}=3.72\, ^\circ C}$$;   Tf   =  0 - ΔTf  =  0 - 3.72°C  = -3.72°C

8.  ΔTf  = 1.86 °C x moles solute particles/kg water ;  calcium chloride = CaCl2

$$\mathrm{moles\: particles = 27.8\: g\: CaCl_2 \times\dfrac{1\:mol\:CaCl_2}{111\:g\:CaCl_2}\times\dfrac{3\:mol\:ions}{1\:mol\:CaCl_2}=0.751\: mol\: particles}$$

$$\mathrm{\Delta T_f=1.86\,^\circ C\times\dfrac{0.751\:mol\:particles}{0.250\:kg\:H_2O}=5.59\,^\circ C}$$;         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

$$\mathrm{moles=\dfrac{\Delta T_f}{1.86}\times kg\:H_2O=\dfrac{4.65\, ^\circ C}{1.86}\times 0.200=0.500\:moles}$$

$$\mathrm{molar\: mass =\dfrac{80.0\: g}{0.500\: moles}= 160\: \dfrac{g}{mole}}$$

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

$$\mathrm{\Delta T_b=2.53\,^\circ C/m\times\dfrac{0.500\:mol}{0.200\:kg}=6.32\,^\circ C}$$;  Tb  =  80.1°C + 6.32°C  = 86.4 °C

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

$$\mathrm{moles=\dfrac{\Delta T_f}{2.53}\times kg\:benzene =\dfrac{10.1\,^\circ C}{2.53}\times0.0750=0.300\:moles}$$

$$\mathrm{molar\: mass=\dfrac{45.0\:g}{0.300\:moles}=150\: \dfrac{g}{mole}}$$