# Answers to More Chapter 11 Study Questions

1. 5.00 g $$\ce{NaClO}$$ per 100. g solution; contains 95.0 g $$\ce{H2O}$$.

$$\mathrm{5.00\:g\:NaClO\times\dfrac{1\:mole\:NaClO}{74.5\:g\:NaClO}=0.0671\:mol}$$; $$\mathrm{95.0\:g\:H_2O\times\dfrac{1\:kg\:H_2O}{1000\:g\:H_2O}=0.0950\:kg\:H_2O}$$

$$\mathrm{molality=\dfrac{moles\:NaClO}{kg\:H_2O}=\dfrac{0.0671\:moles\:NaClO}{0.0950\:kg\:H_2O}=0.706\: m}$$

$$\mathrm{95.0\:g\:H_2O\times\dfrac{1\:mole\:H_2O}{18.0\:g\:H_2O}=5.28\:moles\:H_2O}$$;  $$\mathrm{total\: moles = 5.28 + 0.07 = 5.35\: moles}$$

$$\mathrm{mole\:fraction=\dfrac{moles\:NaClO}{total\:moles}=\dfrac{0.0671\:moles\:NaClO}{5.35\:moles}=0.0125}$$

1. $$\mathrm{12.8\:g\:NaOH\times\dfrac{100\:g\:solution}{32.0\:g\:NaOH}=40.0\:g\:solution}$$

1. 0.0476 moles $$\ce{NaOH}$$; $$\mathrm{(1-0.0476) = 0.952\: mole\: H_2O}$$.

$$\mathrm{0.0476\:mol\:NaOH\times\dfrac{40.0\:g\:NaOH}{1\:mol\:NaOH}=1.90\:g\:NaOH}$$

$$\mathrm{0.952\:mol\:H_2O\times\dfrac{18.0\:g\:H_2O}{1\:mol\:H_2O}=17.1\:g\:H_2O}$$;  $$\mathrm{\textrm{mass of solution} = 17.1 + 1.90 = 19.0\: g\: solution}$$

$$\mathrm{mass\:percent=\dfrac{mass\:NaOH}{mass\:solution}\times100\%=\dfrac{1.90\:g\:NaOH}{19.0\:g\:solution}=10.0\%\:NaOH}$$

$$\mathrm{\dfrac{0.0476\:mol\:NaOH}{19.0\:g\:solution}\times\dfrac{1.11\:g}{1\:mL}\times\dfrac{1000\:mL}{1\:L}=2.78\:M\:NaOH}$$

1. A saturated solution at 80.°C contains 167 g $$\ce{KNO3}$$ dissolved in 100. g $$\ce{H2O}$$.

$$\mathrm{167\:g\:KNO_3\times\dfrac{1\:mol\:KNO_3}{101.1\:g\:KNO_3}=1.65\:mol\:KNO_3}$$

$$\mathrm{100.\:g\:H_2O\times\dfrac{1\:mol\:H_2O}{18.0\:g\:H_2O}=5.56\:mol\:H_2O}$$;  $$\mathrm{total\: moles = 1.65 + 5.56 = 7.21\: moles}$$

$$\mathrm{mole\:fraction=\dfrac{moles\:KNO_3}{total\:moles}=\dfrac{1.65\:moles\:KNO_3}{7.21\:moles}=0.229}$$

1. 44.6 g $$\ce{C7H8}$$ in 250. g of benzene.

From the chart, $$\mathrm{bp(benzene) = 80.10^\circ C}$$;  $$\mathrm{kb(benzene) = 2.53\, ^\circ C\cdot kg/mol}$$

$$\mathrm{\Delta T_b=k_b\times\dfrac{mol\:solute\:particles}{kg\:solvent}}$$

$$\mathrm{44.6\:C_7H_8\times\dfrac{1\:mol\:C_7H_8}{92.0\:g\:C_7H_8}=0.485\:moles\:C_7H_8}$$

$$\mathrm{250.\:g\:benzene\times\dfrac{1\:kg\:benzene}{1000\:g\:benzene}=0.250\:kg\:benzene}$$

$$\mathrm{\Delta T_b=2.53\:^\circ C\times\dfrac{0.485\:moles}{0.250\:kg\:benzene}=4.91\:^\circ C}$$;   $$\mathrm{T_b = 80.10 + 4.91 = 85.01^\circ C}$$

1. $$\mathrm{\Delta T_F=k_F\times\dfrac{mol\:solute\:particles}{kg\:solvent}}$$;  $$\mathrm{k_F(H_2O) = 1.86\, ^\circ C\cdot kg/mol}$$; $$\mathrm{\Delta T_f = 10.0^\circ C}$$

$$\mathrm{\Delta T_F=1.86\times\dfrac{moles\:solute\:particles}{kg\:H_2O}}$$; $$\mathrm{10.0\:^\circ C=1.86\times\dfrac{moles\:solute\:particles}{0.200\:kg\:H_2O}}$$

$$\mathrm{moles=\dfrac{10.0}{1.86}\times0.200\:kg=1.08\:mol\:particles}$$

$$\mathrm{1.08\:mol\:particles\times\dfrac{1\:mol\:CaCl_2}{3\:mol\:particles}\times\dfrac{111\:g\:CaCl_2}{1\:mol\:CaCl_2}=40.0\:g\:CaCl_2}$$

1. 9.00 g solute; 300.0 g water; $$\mathrm{T_f = -0.930^\circ C}$$.  $$\mathrm{\Delta T_F=k_F\times\dfrac{mol\:solute\:particles}{kg\:solvent}}$$

$$\mathrm{300.\: g\: H_2O = 0.300\: kg\: H_2O}$$.  $$\mathrm{\Delta T_f = 0.930^\circ C}$$;  $$\mathrm{k_F(H_2O) = 1.86\, ^\circ C\cdot kg/mol}$$

$$\mathrm{0.930\,^\circ C=1.86\,\dfrac{^\circ C\cdot kg}{mol}\times\dfrac{mol\:solute}{0.300\:kg\:H_2O}}$$;  $$\mathrm{moles=\dfrac{0.930}{1.86}\times0.300=0.150\:moles}$$

$$\mathrm{molar\:mass=\dfrac{mass}{moles}=\dfrac{9.00\:g}{0.150\:mol}=60.0\:g/mol}$$;  alcohol is propanol:  $$\ce{C3H8O}$$