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122ch13f

# 122ch13f - X A = mole fraction of A in the solution P o A =...

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58 13.81) a) b)

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59 13.84)
60 13.87)

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61 13.93)
62 13.93) (cont.)

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63 13.94) Note : To determine the moles of C 2 H 5 OH we could also use the fact that the ratio of mole fractions of two substances is the same as their ratio of moles. X C2H5OH mol C 2 H 5 OH ------------ = ------------------ X C24H50 mol C 24 H 50 X C2H5OH mol C 2 H 5 OH = ------------ (mol C 24 H 50 ) X C24H50 = (0.08/0.92) (1.83 x 10 3 mol) = 159.13 mol C 2 H 5 OH
64 13.96) Given the vapor pressure of pure water and an aqueous NaCl solution and moles of H 2 O find the moles of NaCl. Can use the Raoult’s Law or VP lowering eqn to determine the mole fraction of H 2 O and then use the definition of mole fraction to calculate moles of NaCl particles (ions). Once we have the moles of NaCl particles we can determine the moles of NaCl knowing that it dissociates into 2 ions (particles), i = 2. Raoult’s Law: P A = X A P o A for a volatile substance, A where, P A = vapor pressure of A above the solution

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Unformatted text preview: X A = mole fraction of A in the solution P o A = vapor pressure of pure A For a nonvolatile solute (such as NaCl) the pressure above the solution is due only to the volatile solvent, in this case H 2 O. P solution = P H2O = X H2O P o H2O H 2 O is volatile so the vapor above the solution is just water vapor. P H2O 25.7 torr X H2O = --------- = -------------- = 0.80 8 1 P o H2O 31.8 torr mol H 2 O 0.115 mol H 2 O X H2O = ---------------------------- = ----------------------------------- = 0.80 8 1 (mol ions + mol H 2 O) (mol ions + 0.115 mol H 2 O) 0.115 mol H 2 O = (mol ions + 0.115 mol H 2 O) * 0.80 8 1 mol ions = 0.027 2 95 mol ions mol ions = i * mol NaCl = 2 * mol NaCl mol NaCl = mol ions/2 = (0.027 2 95 mol ions)/2 = 0.013 6 4 = 0.0136 66 13.99) 67 13.100) 68 13.100) (cont.)...
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122ch13f - X A = mole fraction of A in the solution P o A =...

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