Departamento de Química
2003-2004
Fundamentos de Química I (Lic. Química e Bioquímica)
12.3
SF4
Polar (+ solúvel em água )
IF5
Polar (+ solúvel em água)
A X5E
A X4E
SF6
Apolar (− solúvel em água)
AsF5
Apolar (− solúvel em água)
A X6
A X5
12.8 (a) R = −OH ⇒ molécula hidrofílica
(b) R = −CH2 CH3 ⇒ molécula hidrofóbica
(c) R = −CONH2 ⇒ molécula hidrofílica
12.13 (a) S = KH.x p K H (O2 )20°C = 1.3x10-3 mol/L.atm p(O 2 ) = 50. kPa
S = 1.3x10-3 mol/Latm x 50 kPa.atm/101.325 kPa = 6.4x10-4 ml/L
(b) S = KH.x p K H (CO 2 )20°C = 2.3x10-2 mol/L.atm p(CO2 ) = 500. Torr
S = 2.3x10-2 mol/L.atm x 500Torr x 0.1333 kPa.Torr-1 /101.325 kPa.atm-1
S = 1.5x10-2 ml/L
(c) S = 2.3x10-2 mol/L.atm x 0.10 atm = 2.3x10-3 mol/L
12.14 (a) S = KH x.p K H (ar)20°C = 7.9x10-4 mol/L.atm p(ar) = 1 atm
p(N 2 ) = 0.78x 1 p(O 2 ) = 0.22x1
S = 7.9x10-4 mol/Latm x 1 atm = 6.4x10-4 ml/L
(b) S = KH.x p K H (He)20°C = 3.7x10-4 mol/L.atm p(He) = 1 atm
S = 3.7x10-4 mol/L.atm x 1atm x Mr(He) g/mol
S = 3.7x10-4 ml/L x 4.003 g/mol = 1.5 mg/L
Fundamentos de Química I ♣ Cap 12 ♣ 2003-2004
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(c) S = KH.x p K H (He)20°C = 3.7x10-4 mol/L.atm p(He) = 25 kPa
S = 0.37mg/L
12.15 S(CO2 ) = 2.3x10-2 mol/L.atm x 3.00 atm = 6.9x10-2 mol/L
n = 6.9x10-2 mol/L x 0.355 L = 2.45x10-2 mol
V = nRT/P (P = 1.00 atm T = 20°C)
V = 0.59 L
12.6 (a) 4 mg/kg = 4 ppm
(b) p = 0.1923 atm
(c) p = 0.916 atm
12.23+12.24 (a) ∆Hsol = (m/Mr)x ∆H° sol
∆Hsol = 10.0 g / 58.4 g/mol x 3.9 kJ/mol = 0.67 kJ
(Calor absorvido na dissolução do sal = calor libertado pela água no processo = mC S∆T)
0.67x103 J = 100.0 g x 4.18 J/Kg x ∆T
∆T = 1.6 K (Variação de temperatura da água)
NaCl
NaBr
AlCl3
NH4 NO3
KCl
MgBr2
KNO3
NaOH
Mr (g/mol)
58.443
33.801
133.341
80.043
74.551
184.113
101.102
39.997
∆H° sol
∆Hsol
∆T
3.9
-0.6
-329
25.7
17.2
-185.6
34.9
-44.5
0.667
-0.178
-24.67
3.21
2.31
-10.08
3.45
-11.13
1.60
-0.42
-59.03
7.68
5.52
-24.12
8.26
-26.62
12.25
Sr2+(g)
∆HL = 2153 kJ/mol
+
2 Cl− (g)
∆Hh (Sr2+(g))
= -1524 kJ/mol
2 x ∆Hh (Cl−(g))
= -340 kJ/mol
SrCl2 (s)
∆Hsol
Sr2+(aq)
+ 2 Cl− (aq)
∆Hsol (SrCl2 (s)) = ∆HL (SrCl2 (s)) + ∆Hh (Sr2+(g)) + 2 x ∆Hh (Cl− (g))
∆Hsol (SrCl2 (s)) = 1mol x 2153 kJ/mol + 1mol x (−1524 kJ/mol) + 2 mol x (−340 kJ/mol)
∆Hsol (SrCl2 (s)) = − 51 kJ (= − 51 kJ/mol)
Atkin s PW and Jones LL, Chemistry, Molecules, Matter and Change, 4Th Ed., 1999, W H Freeman & Co., ISBN 071672832X
Fundamentos de Química I ♣ Cap 12 ♣ 2003-2004
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12.26
H+(g) + Br−(g) → H+(aq) + Br−(aq)
∆Hh = ∆Hhid (H+(g)) + ∆Hhid (Br−(g))
− 1439 kJ/mol = − 1130 kJ/mol + ∆Hhid (Br−(g))
∆Hhid (Br−(g)) = − 309 kJ/mol
∆Hsol (RbBr(s)) = ∆HL (RbBr(s)) + ∆Hhid (Rb +(g)) + ∆Hhid (Br−(g))
22kJ/mol = 651 kJ/mol + ∆Hhid (Rb +(g)) − 309 kJ/mol
∆Hhid (Rb +(g)) = − 320 kJ/mol
12.45
P = χsolvente Ppuro
(a) χsacarose = 0.100 ⇒ χágua = 1− χsacarose = 0.900
P(H2 O)100ºC = 760 Torr
P = 0.900 x 760 = 684Torr
(b) molalidade = moles de soluto/massa de solvente (= n soluto/kg solvente)
χsoluto = nsoluto /(nsolvente + nsoluto )
nsoluto = m
nsolvente = 1000 g / Msolvente g/mol
χsoluto = nsoluto / [(1000/Msolvente) + nsoluto ]
nsoluto = m = 0.100 mol/kg
nsolvente = 1000 g / 18.02 g/mol = 55.49 mol
χsoluto = 0.100 / [55.49 + 0.100]
χsolvente = 1 - χsoluto
P = 0.9982 x 760 = 759 Torr
12.45
(a) P = (1 - χsoluto ) Ppuro
P = (1-0.050)x355.26 Torr
P =338 Torr
(b) P = {1 – [0.10 / (55.49 + 0.10)]} x 23.76
P = 23.7
Atkin s PW and Jones LL, Chemistry, Molecules, Matter and Change, 4Th Ed., 1999, W H Freeman & Co., ISBN 071672832X
Fundamentos de Química I ♣ Cap 12 ♣ 2003-2004
4/4
12.53
(a) ∆T = kb x m
∆T = 0.51 Kkg/mol x 0.10 mol/kg =0.051 K
pe = 100.051 ºC
(b) ∆T = i x kb x m
∆T = 2 x 0.51 x 0.22 = 0.22
pe = 100.22 ºC
(c) m = 0.230g/0.1kg = 0.08867 molal
∆T = i x kb x m
∆T = 2 x 0.51 x 0.08867 = 0.090
pe = 100.090 ºC
12.56
∆T = kb x m
n = 2.24 g / M g/mol
m = n mol / 0.15 kg = (15 / M) molal
k = 2.79 Kkg/mol
∆T = 0.481 K
0.481 = 2.79 x (15 / M)
M = 87.0 g/mol
12.63
(a) ∆T = i x kf x m
1% NaCl (m/m) ⇒ 1 g NaCl +99 g solvente = 100g solução
m = (1 g / 58.44 g/mol) / 0.099 kg = 0.173 molal
0.593 = i x1.86 x 0.173
i = 1.84
(b) % ionização = (i -1)x100 = 84%
12.70
Π = i R TM
M = molaridade (mol/l)
M = massa molar (g/mol)
(a) Π = 2.30 Torr = 2.30 x 133.3 Pa = 0.3066 kPa
i=1
0.3066 kPa = 8.3145 L.kPa K-1 mol-1 x 293 K x 0.1 g /(M g/mol x 0.2 L)
M = 3973 g/mol
12.73
Π = i RT M
(a) Π = 1 RT M = 2.44x103 kPa/mol x 0.050 mol/l = 122 kPa = 1.2 atm
(b) Π = 2 RT M = 2 x 2.44x103 kPa/mol x 0.0010 mol/l = 4.88 kPa = 0.048 atm
(c) Π = 2 RT M = 2 x 2.44x103 kPa/mol x 2.3x10-5 /133.89g/mol/ 0.100 L (100g=100
mL)
Π = 2 x 2.44x103 kPa/mol x 1.718x10-6 mol/l = 8.38x10-3 kPa = 8.3x10-5 atm
Atkin s PW and Jones LL, Chemistry, Molecules, Matter and Change, 4Th Ed., 1999, W H Freeman & Co., ISBN 071672832X