Exercício Resolvido - Quimica Geral (117)

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Chapter 6: Gases 
 233 
70. (M) We first determine the “dry gas” pressure of helium. This pressure, subtracted 
from the barometric pressure of 738.6 mmHg, gives the vapor pressure of hexane at 
25 C . 
 P
nRT
V
F
HG
I
KJ1072
1
0 08206 298 2
8 446
760
589 7
. . .
.
.
 g
 mol He
4.003 g He
L atm
mol K
K
 L
 mmHg
1 atm
 mmHg 
vapor pressure of hexane = 738.6 589.7 = 148.9 mmHg 
 
71. (M) The first step is to balance the equation: 
 
 2NaClO3 2NaCl + 3O2 
 
 The pressure of O2 is determined by subtracting the known vapor pressure of water at the given 
temperature from the measured total pressure. 
 
 PO2 = PTOT – PH2O = 734 torr – 21.07 torr = 713 torr 
 Patm = 713 mmHg / 760 mmHg = 0.938 atm 
 
2 1
3
3
2 3
2 3
3
0.938 atm 0.0572 LPV
mol O = = = 0.00221 mol
RT 0.08206 L atm K (296 K)
Mass of NaClO is then determined as follows:
2 mol NaClO 106.44 g
0.00221 mol O = 0.1568 g NaClO
3 mol O 1 mol NaClO
%NaClO
0.1568 g
 = 100 17.9%
0.8765 g
 
 
72. (M) The work for this problem is nearly identical to the above problem. 
 
 2KClO3 2KCl + 3O2 
 
 The pressure of O2 is determined by subtracting the known vapor pressure of water at the given 
temperature from the measured total pressure. 
 
 PO2 = PTOT – PH2O = 323 torr – 25.22 torr = 298 torr 
 
Patm = 298 mmHg / 760 mmHg = 0.392 atm 
 
Chapter 6: Gases 
 
234 
2 1
3
3
2 3
2 3
3
0.392 atm 0.229 LPV
mol O = = = 0.003657 mol
RT 0.08206 L atm K (299 K)
Mass of KClO is then determined as follows:
2 mol KClO 122.54 g
0.003657 mol O = 0.299 g KClO
3 mol O 1 mol KClO
0.
%KClO = 
299 g
 100 72%
0.415 g
 
Kinetic-Molecular Theory 
 
73. (M) Recall that 1 J = kg m
2
 s
-2 
2 -2
rms 3
2
2
J 1 kg m s
3 8.3145 303 K
mol K 1 J3
326 m/s
70.91 10 kg Cl
1 mol Cl
RT
u
M
 
 
74. (M) 
1
1 1
1 1
22
3
2 = Square both sides and solve for . = 4 273 K = 1092 K
273 K3
RT
T TM
T T
TRT
M
 
Alternatively, recall that 1 J = kg m
2
 s
-2 
u
RT
M
rms m/s
3
184 103. Solve this equation for temperature with urms doubled. 
2 3 3 2
3rms
2 -2
2.016 10 kg/mol (2 1.84 10 m /s)
1.09 10 K
3 J 1 kg m s
3 8.3145
mol K 1 J
Mu
T
R
 
 
75. (M) 
2 2
rms
J
3 8.3145 298K
3 mol K 0.00783 kg / mol = 7.83 g/mol. or 7.83 .
( ) mi 1 hr 5280 ft 12 in. 1 m
2180
hr 3600 sec 1 mi 1 ft 39.37 in.
RT
M u
u
 
 
76. (E) A noble gas with molecules having urms at 25 C greater than that of a rifle bullet will 
have a molar mass less than 7.8 g/mol. Helium is the only possibility. A noble gas with a 
slower urms will have a molar mass greater than 7.8 g/mol; any one of the other noble gases 
will have a slower urms . 
 
77. (E) We equate the two expressions for root mean square speed, cancel the common 
factors, and solve for the temperature of Ne. Note that the units of molar masses do not