Answers to Chapter 11 Study Questions
- Page ID
- 11281
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- \(\mathrm{sodium\: chlorate = NaClO_3}\)
\(\mathrm{284\: g\: solution \times \dfrac{12.0\:g\:NaClO_3}{100\:g\:solution}\times\dfrac{1\:mol\:NaClO_3}{106.4\:g\:NaClO_3}=0.320\: moles}\)
- \(\mathrm{mole\: fraction =\dfrac{moles\:C_2H_6O_2}{total\:moles}}\); \(\mathrm{moles\: C_2H_6O_2 = 120.0\: g\times\dfrac{1\:mol\:C_2H_6O_2}{62.0\:g\:C_2H_6O_2}=1.94\: moles}\)
\(\mathrm{moles\: C_3H_6O = 1.20\: kg\times\dfrac{1000\:g}{1\:kg}\times\dfrac{1\:mol\:C_3H_6O}{58.0\:g\:C_3H_6O}=20.7\: moles}\)
\(\mathrm{total\: moles = 1.94 + 20.7 = 22.6\: moles}\); \(\mathrm{mole\: fraction =\dfrac{1.94\:mol\:C_2H_6O_2}{22.6\:moles}=0.0857}\)
- \(\mathrm{\dfrac{6.90\:mol\:KOH}{1\:L\:solution}\times\dfrac{1\:L}{1000\:mL}\times\dfrac{1\:mL}{1.29\:g}\times\dfrac{56.1\:g\:KOH}{1\:mol\:KOH}=\dfrac{0.300\:g\:KOH}{g\:solution}=30.0\%\: KOH}\)
\(\mathrm{30.0\%\: KOH = 30.0\: g\: KOH + 70.0\: g\: H_2O;\: molality = \dfrac{moles\: KOH}{kg\: water}}\)
\(\mathrm{\dfrac{30.0\:g\:KOH}{70.0\:g\:H_2O}\times\dfrac{1000\:g\:H_2O}{1\:kg\:H_2O}\times\dfrac{1\:mol\:KOH}{56.1\:g\:KOH}=\dfrac{7.64\: moles\: KOH}{kg\: water}}\)
- \(\mathrm{methanol = CH_3OH}\)
\(\mathrm{molarity=\dfrac{moles\:CH_3OH}{L\:solution}}\); \(\mathrm{12.8\: g\: CH_3OH\times\dfrac{1\:mol\:CH_3OH}{32.0\:g\:CH_3OH}=0.400\: mol\: CH_3OH}\)
\(\mathrm{volume\: CH_3OH = 12.8\: g\times\dfrac{1\:mL}{0.791\:g}=16.2\: mL}\); \(\mathrm{volume\: water = 144\: mL}\)
\(\mathrm{total\: volume\: of\: solution = 16.2\: mL + 144\: mL = 160\: mL = 0.160\: L}\)
\(\mathrm{molarity =\dfrac{0.400\:mol\:CH_3OH}{0.160\:L\:solution}=2.50\:M}\)
- 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 \(\ce{CO2}\) 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.
- 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.
- \(\mathrm{\Delta T_f = 1.86\: ^\circ C \times \dfrac{moles\: solute\: particles}{kg\: water}}\)
\(\mathrm{\Delta T_f = 1.86\, ^\circ C \times\dfrac{0.11\:moles}{0.055\:kg\:H_2O}=3.72\, ^\circ C}\); \(\mathrm{T_f = 0 - \Delta T_f = 0 - 3.72^\circ C = -3.72^\circ C}\)
- \(\mathrm{\Delta T_f = 1.86\: ^\circ C \times \dfrac{moles\: solute\: particles}{kg\: water}}\); \(\mathrm{calcium\: chloride = CaCl_2}\)
\(\mathrm{moles\: particles = 27.8\: g\: CaCl_2 \times\dfrac{1\:mol\:CaCl_2}{111\:g\:CaCl_2}\times\dfrac{3\:mol\:ions}{1\:mol\:CaCl_2}=0.751\: mol\: particles}\)
\(\mathrm{\Delta T_f=1.86\,^\circ C\times\dfrac{0.751\:mol\:particles}{0.250\:kg\:H_2O}=5.59\,^\circ C}\); \(\mathrm{T_f = -5.59\: ^\circ C}\)
(\(\ce{CaCl2}\) is an electrolyte and it is important to remember that there are 3 moles of ions per mole of \(\ce{CaCl2}\). The freezing point is three times lower than it would be for a nonelectrolyte.)
- \(\mathrm{molar\: mass = \dfrac{mass}{moles}}\); \(\mathrm{mass = 80.0\: g}\); find moles
\(\mathrm{moles=\dfrac{\Delta T_f}{1.86}\times kg\:H_2O=\dfrac{4.65\, ^\circ C}{1.86}\times 0.200=0.500\:moles}\)
\(\mathrm{molar\: mass =\dfrac{80.0\: g}{0.500\: moles}= 160\: \dfrac{g}{mole}}\)
- From Table 11.5, for benzene: \(\mathrm{K_b = 2.53\dfrac{^\circ C}{m}}\) and \(\mathrm{T_b = 80.1^\circ C}\).
\(\mathrm{\Delta T_b=2.53\,^\circ C/m\times\dfrac{0.500\:mol}{0.200\:kg}=6.32\,^\circ C}\); \(\mathrm{T_b = 80.1^\circ C + 6.32^\circ C = 86.4 ^\circ C}\)
- \(\mathrm{molar\: mass = \dfrac{mass}{moles}}\); \(\mathrm{mass = 45.0\: g}\); find moles; \(\mathrm{\Delta T_b = 90.2^\circ C - 80.1^\circ C = 10.1^\circ C}\)
\(\mathrm{moles=\dfrac{\Delta T_f}{2.53}\times kg\:benzene =\dfrac{10.1\,^\circ C}{2.53}\times0.0750=0.300\:moles}\)
\(\mathrm{molar\: mass=\dfrac{45.0\:g}{0.300\:moles}=150\: \dfrac{g}{mole}}\)