Exercício Resolvido - Quimica Geral (66)

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Chapter 4: Chemical Reactions 
 131 
 
36. (E) 
 (a) 
3 2
3 2 3 2
CH CH OH
3 2
1.65 mol CH CH OH 46.07 g CH CH OH 1mL
V = 200.0 L soln
1L 1mol CH CH OH 0.789 g
 
 
1L
= 19.3L
1000 mL
 (b)
 HCl
0.234mol HCl 36.46g HCl 100g soln 1mL soln
V = 12.0L
1L 1mol HCl 36.0g HCl 1.18g
= 241mL 
 
37. (M) 
 (a) 6 12 6 6 12 6 6 12 6
-3
6 12 6 6 12 6
85 mg C H O 1 mol C H O 1 mmol C H O1g 10 dL
1 dL blood 1000 mg 1 L 180.16 g C H O 1 10 mol C H O
 
 
6 12 6mmol C H O= 4.7 
L
 
 (b) 3 6 12 6
 mol C H O
Molarity 4.7 10
L
 
 
38. (M) (a) 5
1.2 mg F 1g 1 mol F mol F
Molarity = 6.3 10 
1 L 1000 mg 19.00 g F L
 
 
 (b) 
5 -
8 56.3 10 mol F 1 mol KF 58.1g KF# g KF 1.6 10 L water 5.9 10 g KF
1 L water 1 mol F 1 mol KF
 
 
39. (E) First we determine each concentration in moles per liter and find the 0.500 M solution. 
 (a) 
0.500 g KCl 1 mol KCl 1000 mL
[KCl] = = 6.71 M KCl
1 mL 74.551 g KCl 1 L
 
 (b) 
36.0 g KCl 1 mol KCl
[KCl] = = 0.483 M KCl
1 L 74.551 g KCl 
 
 (c) 
7.46 mg KCl 1 g KCl 1 mol KCl 1000 mL
[KCl] = = 0.100 M KCl
1 mL 1000 mg KCl 74.551 g KCl 1 L
 
 (d) 
373 g KCl 1 mol KCl
[KCl] = = 0.500 M KCl
10.00 L 74.551 g KCl 
 
 Solution (d) is a 0.500 M KCl solution. 
 
40. (E) By inspection, we see that (b) and (c) are the only two that are not per volume of 
solution. These two solutions need not be considered. A close inspection of the remaining 
choices reveals that the units for (a) are equivalent to those for (d), that is g NaCl per liter of 
solution is equivalent to mg NaCl per mL of solution (the mass:volume ratio is the same). 
 
 
 
Chapter 4: Chemical Reactions 
132 
 
41. (E) We determine the molar concentration for the 46% by mass sucrose solution. 
 
12 22 11
12 22 11
12 22 11
12 22 11
1 molC H O
46 g C H O ×
342.3 g C H O
[C H O ] = = 1.6 M
1 mL 1 L
100 g soln× ×
1.21 g soln 1000 mL
 
 The 46% by mass sucrose solution is the more concentrated. 
 
42. (E) Here we must calculate the 3 2CH CH OH in the white wine and compare it with 
 1.71 M 3 2CH CH OH , the concentration of the solution described in Example 4-8. 
 
3 2 3 2
3 2
3 2
3 2
11 g CH CH OH 1 mol CH CH OH0.95 g soln 1000 mL
CH CH OH =
100.0 g soln 1 mL 1 L 46.1g CH CH OH
= 2.3 M CH CH OH
 
 Thus, the white wine has a greater ethyl alcohol content. 
 
43. (E) 
3
3
2.05mol KNO
0.01000Lconc'd soln
1L
[KNO ] = = 0.0820M
0.250 L diluted solution
 
 
44. (E) 
 
3
3
3
AgNO
3
Volume of concentrated AgNO solution
0.425 mmol AgNO 1 mL conc. soln.
V = 250.0 mL dilute soln × × = 142 mL conc. soln
1 mL dilute soln 0.750 mmol AgNO
 
 
45. (E) Both the diluted and concentrated solutions contain the same number of moles of K SO2 4 . This 
number is given in the numerator of the following expression. 
2 4
2 4
0.198 mol K SO
0.125 L
1L
K SO molarity = = 0.236 M
0.105 L
K2SO4 
 
46. (E) 
0.085mol HCl
0.500 Ldilutesol'n
1Lsoln
[HCl] = = 1.7 M
0.0250L
 
 
47. (E) Let us compute how many mL of dilute (d) solution we obtain from each mL of 
concentrated (c) solution. V C V Cc c d d= becomes 1.00 mL 0.250M = mL 0.0125x M 
and x = 20 Thus, the ratio of the volume of the volumetric flask to that of the pipet would 
be 20:1. We could use a 100.0-mL flask and a 5.00-mL pipet, a 1000.0-mL flask and a 
50.00-mL pipet, or a 500.0-mL flask and a 25.00-mL pipet. There are many combinations 
that could be used.