*V-1
I-3
I-4
C
C
C
NC
NC
NC
**V-2
II-2
II-3
Quiz 6
NAME:
C
C
C
NC
NC
NC
***V-3
VII-1
VII-2
C
C
C
NC
NC
NC
CHEMISTRY 131-01
Summer 2011 Form B
Key
Calculate molar masses to 0.01 g/mol. Show work for all calculations, and place your final
answer in the space provided. Be sure to give units as well as numbers in the answer, as well
as in calculations.
Chapter 13: Solutions
A.
(1 pt) Put an X or check mark in the space to the left of the substance if it is a solution. If
the substance is not a solution, leave the space blank.
X
Mountain Dew soft drink
House paint
Cream
Distilled water
X
Sea water
*B. (2 pts) 45.3 grams of Na2S are dissolved in enough water to make 2.50 L of solution.
1.
What is the molarity of the solution?
Answer:
0.232 M Na2S
45.3 g Na2S x 1 mol Na2S
2.50 L sol'n
78.05 g Na2S
2.
= 0.304 mol Na2S = 0.232 M Na2S
L sol'n
What is the mass percent Na2S in the solution? Its density is 1.03 g/mL.
Answer:
1.76 % by mass Na2S
45.3 g Na2S
x 1 mL sol'n x 100% = 1.76 % by mass Na2S
2.50 x 103 mL sol'n
1.03 g sol'n
C.
You have been asked to prepare 250 mL of 0.500 M Na2CO3.
*1.
(0.7 pt) How much Na2CO3 should you use?
Answer:
13.2 g Na2CO3
0.500 mol Na2CO3 x 105.99 g Na2CO3 x 0.250 L sol’n = 13.2 g Na2CO3
L sol'n
mol Na2CO3
2.
(0.3 pt) What is the correct way to prepare the solution? Check the box with the
correct response.
Put the Na2CO3 in a vessel and add 250 mL of water.
X
Put the Na2CO3 in a vessel and dilute to 250 mL with water.
Put the Na2CO3 in a vessel and add 237 mL of water.
D.
You have been asked to prepare 500 mL of 20.0% (v/v) ethanol (C2H6O).
*1.
(0.7 pt) How much C2H6O should you use?
Answer:
100 mL C2H6O
20.0 mL C2H6O x 500 mL sol'n = 100 mL C2H6O
100 mL sol'n
2.
(0.3 pt) What is the correct way to prepare the solution? Check the box with the
correct response.
X
Put 100 mL C2H6O in a vessel and dilute to 500.0 mL with water.
Put 100 ml C2H6O in a vessel and add 400.0 mL of water.
Put 100 g C2H6O in a vessel and add 400.0 g of water.
**E. (2 pts) Your instructor must prepare 250 mL of 10.0% (m/m) formalin (CH2O) for an
experiment. Concentrated formalin is 37.0% (m/m), and its density is 1.08 g/mL.
1.
What volume of concentrated formalin should you use?
Answer:
67.6 mL
C1 = 37.0% (m/m)
V1 =
x
C2 = 10.0% (m/m)
V2 = 500 mL
2.
C1 V1 = C2 V2
V1 = C2 V2 = 10.0% (m/m) x 250 mL = 67.6 mL
C1
37.0% (m/m)
What is the molarity of concentrated formalin?
Answer:
13.3 M
37.0 g CH2O x 1 mol CH2O x 1.08 g sol’n x 1 mL sol’n = 13.3 mol
100 g sol’n
30.03 g CH2O
mL sol’n
0.001 L sol’n
L
F.
(2 pts) 25.00 mL of nitrilotriacetic acid, (NTA, C6H9NO6) is titrated with 20.23 mL of
0.09885 M NaOH. The equation for the reaction is shown below.
C6H9NO6 + 3 NaOH
1.
→
Na3C6H6NO6 + 3 H2O
What is the molarity of the NTA?
Answer:
0.02666 M
20.23 mL NaOH x 0.09885 mol NaOH x 1 mol NTA x
1
= 0.02666 M
1000 mL NaOH
3 mol NaOH
0.02500 L MA
2.
How many grams of NTA are present in the sample?
Answer:
0.1274 g NTA
0.02666 mol MA x 0.02500 L MA x 191.14 g NTA = 0.1274 g NTA
L MA
mol NTA
***G.
(2 pts) Choose the correct response for each item below and mark it with an “X” in
the space provided.
1.
Two solutions are separated by a semipermeable membrane through which only
water can pass. One of the solutions is 6.0 M NaF ; the other is 0.10 M NaF. Which
of the following statements is true?
A. NaF from the 0.10 M solution will flow to the 6.0 M solution.
B. Water from the 6.0 M solution will flow to the 0.10 M solution.
X
C. Water from the 0.10 M solution will flow to the 6.0 M solution.
D. NaF from the 6.0 M solution will flow to the 0.10 M solution.
2.
Which of the following statements about reverse osmosis is true?
X
A. Pressure forces solvent through the membrane away from the solution.
B. There is no such thing as reverse osmosis.
C. Solvent is pulled from the concentrated solution to the dilute solution.
D. Solute flows from the concentrated solution to the dilute solution.
Competencies from Previous Quizzes:
I-3
Perchloroethylene (PERC) has been used as a dry-cleaning fluid and degreasing solvent.
It is now thought to be carcinogenic and is not widely use. Its composition is 85.51% Cl
and 14.49 % C. What is its empirical formula? Show your work.
Answer:
CCl2
85.51 g Cl x 1 mol Cl = 2.412 mol Cl
35.45 g Cl
14.49 g C x
I-4
2.412 mol Cl = 2.000 mol Cl = 2 Cl
1.206 mol C
mol C
1 C
1 mol C = 1.206 mol C
12.01 g C
CCl2
Phloroglucinol is a stain used for microscopy of plant cells. It imparts a red color to lignin,
which binds cellulose in woody plants. Its empirical formula is C2H2O, and its molar mass
is 126.11 g/mol. What is its molecular formula? Show your work.
Answer:
C6H6O3
2 mol C x 12.01 g C = 24.02 g C
mol C
2 mol H x
1.01 g H =
mol H
2.02 g H
1 mol O x 16.00 g O = 16.00 g O
mol O
42.04 g/formula
126.11 g x 1 formula = 3.000 formula
mol
42.04 g
mol
3(C2H2O) = C6H6O3
II-2 Sodium azide, NaN3, releases nitrogen gas in automotive airbags and other applications.
The reaction equation is
2 NaN3
→
2 Na
+
3 N2
What mass of N2 would be released by decomposition of 5.00 g of NaN3? Show your
work.
Answer:
3.23 g N2
5.00 g NaN3 x 1 mol NaN3 x 3 mol N2 x 28.02 g N2 = 3.23 g N2
65.02 g NaN3
2 mol NaN3
mol N2
II-3 A chemist has prepared a new compound that has promise as an analgesic. She obtained 25.459 g of the compound, but she had calculated her theoretical yield to be
83.568 g. What was her percent yield? Show your work?
Answer:
30.465%
25.459 g x 100% = 30.465%
83.568 g
VII-1 There is 1.00 mol of Cl2 gas in a small container with a volume of 515 mL. If its temperature is 25°C, what is the pressure of the gas in the bottle? Show your work.
Answer:
47.5 atm
P = x
V = 515 mL x 0.001 L = 0.515 L
mL
n = 1.00 mol
T = 25°C + 273 = 298 K
PV = nRT
P = nRT
V
= 1.00 mol x 298 K x 0.0821 L ⋅ atm
0.515 L
mol ⋅ K
= 47.5 atm
VII-2 Consider the following molecules, then answer the questions below.
1.
Cl−Cl
Br−Br
A. Chlorine
B. Bromine
Give the name of the compound that will have the higher boiling point.
Bromine
2.
What forces are stronger in that molecule than in the other one?
Dispersion forces