Deck 12: Solutions
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Deck 12: Solutions
1
A solution is made by mixing 50 grams of CHCl3 (liquid) with 10 grams of CH2Cl2 (liquid). Which of the two is considered the solute and why?
A) CHCl3 is the solute because it is in the greater amount.
B) CHCl3 is the solute because it is in the lesser amount.
C) CH2Cl2 is the solute because it is in the greater amount.
D) CH2Cl2 is the solute because it is in the lesser amount.
E) Neither is the solute because they are both liquids at ambient conditions and therefore are solvents.
A) CHCl3 is the solute because it is in the greater amount.
B) CHCl3 is the solute because it is in the lesser amount.
C) CH2Cl2 is the solute because it is in the greater amount.
D) CH2Cl2 is the solute because it is in the lesser amount.
E) Neither is the solute because they are both liquids at ambient conditions and therefore are solvents.
CH2Cl2 is the solute because it is in the lesser amount.
2
How many milliliters of water must be added to 75.0 mL of 0.150 M AgNO3 in order to prepare a 0.100 M solution?
(Assume volumes are additive.)
A) 25.0 mL of H2O
B) 38.0 mL of H2O
C) 50.0 mL of H2O
D) 113 mL of H2O
E) 188 mL of H2O
(Assume volumes are additive.)
A) 25.0 mL of H2O
B) 38.0 mL of H2O
C) 50.0 mL of H2O
D) 113 mL of H2O
E) 188 mL of H2O
38.0 mL of H2O
3
What mass of KCl is required to prepare 500 mL of a 2.50 M aqueous solution of KCl?
MM(KCl) = 74.55 g/mol.
A) 0.0168 g KCl
B) 0.0671 g KCl
C) 14.9 g KCl
D) 93.2 g KCl
E) 373 g KCl
MM(KCl) = 74.55 g/mol.
A) 0.0168 g KCl
B) 0.0671 g KCl
C) 14.9 g KCl
D) 93.2 g KCl
E) 373 g KCl
93.2 g KCl
4
Three hundred milliliters (300 mL) of 2.50 M sugar solution is diluted tO600 mL by the addition of pure water. Which of these statements about the solution is correct?
A) The moles of water remain the same.
B) The moles of sugar decrease.
C) The concentration of sugar increases tO5 .0 M.
D) The moles of sugar remain constant.
E) All of these statements are incorrect.
A) The moles of water remain the same.
B) The moles of sugar decrease.
C) The concentration of sugar increases tO5 .0 M.
D) The moles of sugar remain constant.
E) All of these statements are incorrect.
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5
Calculate the molarity of a solution prepared by dissolving 43.2 grams of pentane (C5H12, molar mass = 72.15 g/mol) in enough carbon tetrachloride (CCl4, molar mass = 153.8 g/mol) to take 400 mL of solution.
A) 10.8 M
B) 0.600 M
C) 6.01 M
D) 4.17 M
E) 1.50 M
A) 10.8 M
B) 0.600 M
C) 6.01 M
D) 4.17 M
E) 1.50 M
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6
The molarity and molality of a solution:
A) may never be equal.
B) are equal for all solutions.
C) both express the quantity of solute in moles.
D) are equal for all solutions having water as the solvent.
E) are equal for all concentrated solutions.
A) may never be equal.
B) are equal for all solutions.
C) both express the quantity of solute in moles.
D) are equal for all solutions having water as the solvent.
E) are equal for all concentrated solutions.
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7
What is the definition for mass percent solute ?
A) (grams of solute / grams of solvent)×100%
B) (grams of solvent / grams of solute)×100%
C) (moles of solute / moles of solvent)×100%
D) (grams of solute / grams of solution)×100%
E) (moles of solution / moles of solute)×100%
A) (grams of solute / grams of solvent)×100%
B) (grams of solvent / grams of solute)×100%
C) (moles of solute / moles of solvent)×100%
D) (grams of solute / grams of solution)×100%
E) (moles of solution / moles of solute)×100%
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8
How many grams of silver nitrate (AgNO3) are present in 255 mL of 2.50 M silver nitrate solution?
MM(AgNO3) = 169.87 g/mol
A) 5.77×10 - 2 g AgNO3
B) 17.3 g AgNO3
C) 108 g AgNO3
D) 266 g AgNO3
E) 1.67×103 g AgNO3
MM(AgNO3) = 169.87 g/mol
A) 5.77×10 - 2 g AgNO3
B) 17.3 g AgNO3
C) 108 g AgNO3
D) 266 g AgNO3
E) 1.67×103 g AgNO3
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9
A solution is made by mixing 25 grams of ethanol witH5 grams of isopropanol. Which of the two is considered the solvent and why?
A) Ethanol is the solvent because it is in the greater amount.
B) Ethanol is the solvent because it is in the lesser amount.
C) Isopropanol is the solvent because it is in the greater amount.
D) Isopropanol is the solvent because it is in the lesser amount.
E) Both are solvents because they are both liquids at ambient conditions.
A) Ethanol is the solvent because it is in the greater amount.
B) Ethanol is the solvent because it is in the lesser amount.
C) Isopropanol is the solvent because it is in the greater amount.
D) Isopropanol is the solvent because it is in the lesser amount.
E) Both are solvents because they are both liquids at ambient conditions.
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10
How many grams of H2SO4 are present in 75.0 mL of 12.0 M H2SO4?
MM (H2SO4) = 98.09 g/mol
A) 1.63 g H2SO4
B) 9.18 g H2SO4
C) 15.7 g H2SO4
D) 88.3 g H2SO4
E) 613 g H2SO4
MM (H2SO4) = 98.09 g/mol
A) 1.63 g H2SO4
B) 9.18 g H2SO4
C) 15.7 g H2SO4
D) 88.3 g H2SO4
E) 613 g H2SO4
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11
How many grams of urea [CO(NH2)2] would be required to prepare 200 mL of a 5.0×10 - 3 molar solution?
A) 1.0×10 - 3 g
B) 6.0×10 - 2 g
C) 1.5 g
D) 5.0×10 - 3 g
E) 1.7×10 - 5 g
A) 1.0×10 - 3 g
B) 6.0×10 - 2 g
C) 1.5 g
D) 5.0×10 - 3 g
E) 1.7×10 - 5 g
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12
Potassium fluoride is used for frosting glass. What is the molarity of a solution prepared by dissolving 78.6 g of KF in enough water to produce 225 mL of solution?
MM(KF) = 58.10 g/mol
A) 0.304 M
B) 0.349 M
C) 1.35 M
D) 3.29 M
E) 6.01 M
MM(KF) = 58.10 g/mol
A) 0.304 M
B) 0.349 M
C) 1.35 M
D) 3.29 M
E) 6.01 M
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13
Which of the following concentration terms below represents moles of solute per kilogram of solvent?
A) Molarity
B) Molality
C) Molar mass
D) Mole fraction
E) Mole percent
A) Molarity
B) Molality
C) Molar mass
D) Mole fraction
E) Mole percent
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14
Calculate the molarity of glucose (C6H12O6) in a solution in which 10.0 g of glucose is dissolved in water and diluted to a final volume of 200 mL.
A) 0.278 M
B) 0.0556 M
C) 0.0050 M
D) 90.0 M
E) 50.0 M
A) 0.278 M
B) 0.0556 M
C) 0.0050 M
D) 90.0 M
E) 50.0 M
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15
How many grams of AgNO3 are present in 50.0 mL of a 12.0 M AgNO3 solution?
MM(AgNO3) = 169.88 g/mol
A) 3.53 g of AgNO3
B) 40.8 g of AgNO3
C) 102 g of AgNO3
D) 283 g of AgNO3
E) 708 g of AgNO3
MM(AgNO3) = 169.88 g/mol
A) 3.53 g of AgNO3
B) 40.8 g of AgNO3
C) 102 g of AgNO3
D) 283 g of AgNO3
E) 708 g of AgNO3
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16
What is the definition for molarity ?
A) Grams of solute per moles of solute
B) Moles of solute per grams of solution
C) Moles of solute per Liter of solvent
D) Grams of solute per Liter of solution
E) Moles of solute per Liter of solution
A) Grams of solute per moles of solute
B) Moles of solute per grams of solution
C) Moles of solute per Liter of solvent
D) Grams of solute per Liter of solution
E) Moles of solute per Liter of solution
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17
What is the definition for molality ?
A) kilograms of solute/mole of solvent
B) Avogadro's number of atoms, molecules or formula units
C) moles of solute/Liter of solution
D) moles of solute/Kg of solvent
E) moles of solute/Kg of solution
A) kilograms of solute/mole of solvent
B) Avogadro's number of atoms, molecules or formula units
C) moles of solute/Liter of solution
D) moles of solute/Kg of solvent
E) moles of solute/Kg of solution
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18
How many grams of solute are present in 164.1 mL of 0.207 M calcium acetate, Ca(C2H3O2)2?
MM[Ca(C2H3O2)2] = 158.17 g/mol
A) 0.200 g Ca(C2H3O2)2
B) 0.215 g Ca(C2H3O2)2
C) 4.66 g Ca(C2H3O2)2
D) 5.01 g Ca(C2H3O2)2
E) 5.37 g Ca(C2H3O2)2
MM[Ca(C2H3O2)2] = 158.17 g/mol
A) 0.200 g Ca(C2H3O2)2
B) 0.215 g Ca(C2H3O2)2
C) 4.66 g Ca(C2H3O2)2
D) 5.01 g Ca(C2H3O2)2
E) 5.37 g Ca(C2H3O2)2
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19
How many grams of KBr are required to prepare 1.50 L of 0.257 M aqueous solution?
MM(KBr) = 119.00 g/mol
A) 3.24×10 - 3 g KBr
B) 20.4 g KBr
C) 45.9 g KBr
D) 309 g KBr
E) 695 g KBr
MM(KBr) = 119.00 g/mol
A) 3.24×10 - 3 g KBr
B) 20.4 g KBr
C) 45.9 g KBr
D) 309 g KBr
E) 695 g KBr
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20
What volume in milliliters of a 0.200 M NaCl solution is required to provide 1.00 grams of NaCl?
MM(NaCl) = 58.44 g/mol
A) 3.42 mL
B) 11.7 mL
C) 85.6 mL
D) 292 mL
E) 5000 mL
MM(NaCl) = 58.44 g/mol
A) 3.42 mL
B) 11.7 mL
C) 85.6 mL
D) 292 mL
E) 5000 mL
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21
A solution is prepared by mixing 22 g ethylene glycol [C2H4(OH)2] with 108 g of water. What is the mass percent of ethylene glycol in the solution?
A) 20%
B) 17%
C) 22%
D) 83%
E) none of these
A) 20%
B) 17%
C) 22%
D) 83%
E) none of these
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22
What is the mole fraction of glycerol for a mixture prepared by mixing 300 grams of glycerol (non-volatile solute) and 505 grams of H2O?
Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) c glycerol = 0.104
B) c glycerol = 0.116
C) c glycerol = 0.896
D) c glycerol = 1.12
E) c glycerol = 8.59
Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) c glycerol = 0.104
B) c glycerol = 0.116
C) c glycerol = 0.896
D) c glycerol = 1.12
E) c glycerol = 8.59
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23
How many grams of NaCl are required to prepare 250 mL of 3.60% (m/v) aqueous solution of NaCl?
A) 0.694 g NaCl
B) 1.44 g NaCl
C) 9.00 g NaCl
D) 69.4 g NaCl
E) 900 g NaCl
A) 0.694 g NaCl
B) 1.44 g NaCl
C) 9.00 g NaCl
D) 69.4 g NaCl
E) 900 g NaCl
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24
The mass percent of acetone in a solution prepared by mixing 58.1 g acetone (C3H6O) and 72.1 g of H2O is:
A) 44.6%
B) 80.6%
C) 55.4%
D) 13.9%
E) none of these
A) 44.6%
B) 80.6%
C) 55.4%
D) 13.9%
E) none of these
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25
How many grams of Na2SO4 are required to make 43.5 mL of a 13.7% (m/v) aqueous solution?
A) 0.168 g Na2SO4
B) 5.96 g Na2SO4
C) 13.7 g Na2SO4
D) 31.5 g Na2SO4
E) 318 g Na2SO4
A) 0.168 g Na2SO4
B) 5.96 g Na2SO4
C) 13.7 g Na2SO4
D) 31.5 g Na2SO4
E) 318 g Na2SO4
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26
How many grams of Na2CO3 are required to prepare 25.0 mL of a 2.00% (m/v) Na2CO3 solution?
A) 0.0800 g Na2CO3
B) 0.500 g Na2CO3
C) 2.00 g Na2CO3
D) 12.5 g Na2CO3
E) 50.0 g Na2CO3
A) 0.0800 g Na2CO3
B) 0.500 g Na2CO3
C) 2.00 g Na2CO3
D) 12.5 g Na2CO3
E) 50.0 g Na2CO3
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27
How many grams of glucose (C6H12O6) are required to prepare 50.0 mL of 6.30% (m/v) glucose-water solution?
A) 0.0175 g C6H12O6
B) 0.126 g C6H12O6
C) 3.15 g C6H12O6
D) 6.30 g C6H12O6
E) 7.94 g C6H12O6
A) 0.0175 g C6H12O6
B) 0.126 g C6H12O6
C) 3.15 g C6H12O6
D) 6.30 g C6H12O6
E) 7.94 g C6H12O6
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28
What is the mole fraction of water for a mixture prepared by mixing 50.0 grams of glycerol (non-volatile solute) and 500 grams of H2O?
Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) c water = 0.0192
B) c water = 0.0361
C) c water = 0.0909
D) c water = 0.909
E) c water = 0.981
Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) c water = 0.0192
B) c water = 0.0361
C) c water = 0.0909
D) c water = 0.909
E) c water = 0.981
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29
A solution is prepared by mixing 22 g ethylene glycol [C2H4(OH)2, molar mass = 62 g/mol] with 108 g of water. What is the mole fraction of ethylene glycol in the solution?
A) 0.056
B) 0.17
C) 0.35
D) 6.3
E) none of these
A) 0.056
B) 0.17
C) 0.35
D) 6.3
E) none of these
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30
What is the mass percentage, %(m/m), of an aqueous solution prepared by dissolving 24.5 g of NaCl in 86.5 g of water?
A) 0.283%
B) 3.53%
C) 22.1%
D) 28.3%
E) 77.9%
A) 0.283%
B) 3.53%
C) 22.1%
D) 28.3%
E) 77.9%
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31
Calculate the mole fraction of water in a solution made by mixing 1.0 g of water and 2.0 g of ethyl alcohol (C2H5OH).
A) 75%
B) 67%
C) 33%
D) 56%
E) none of these
A) 75%
B) 67%
C) 33%
D) 56%
E) none of these
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32
A commercial bleaching solution contains 3.62% sodium hypochlorite, NaOCl. What is the mass of NaOCl present in a bottle that contains 4.50×103 g of bleaching solution?
A) 8.04×10 - 4 g NaOCl
B) 0.0804 g NaOCl
C) 163 g NaOCl
D) 1.24×103 g NaOCl
E) 1.63×104 g NaOCl
A) 8.04×10 - 4 g NaOCl
B) 0.0804 g NaOCl
C) 163 g NaOCl
D) 1.24×103 g NaOCl
E) 1.63×104 g NaOCl
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33
What is the mole fraction of glucose in a solution prepared by dissolving 9.01 grams of glucose, C6H12O6, in 17.1 grams of water?
Molar Mass(C6H12O6) = 180.16 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) c glucose = 0.0503
B) c glucose = 0.345
C) c glucose = 0.527
D) c glucose = 0.909
E) c glucose = 0.950
Molar Mass(C6H12O6) = 180.16 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) c glucose = 0.0503
B) c glucose = 0.345
C) c glucose = 0.527
D) c glucose = 0.909
E) c glucose = 0.950
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34
The mole fraction of acetone in a solution prepared by mixing 58.1 g acetone (C3H6O) and 72.1 g of H2O is:
A) 0.200
B) 0.250
C) 0.446
D) 0.806
E) 1.00
A) 0.200
B) 0.250
C) 0.446
D) 0.806
E) 1.00
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35
How many grams of glucose (C6H12O6) must be dissolved in 54 g of water to give a solution in which the mole fraction of glucose is 0.50?
A) 180 g
B) 54 g
C) 27 g
D) 5.4×102 g
E) 0.50 g
A) 180 g
B) 54 g
C) 27 g
D) 5.4×102 g
E) 0.50 g
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36
A solution is prepared by mixing 4.00 grams of benzene (molar mass = 78.1 g/mol), 4.00 grams of CCl4 (molar mass = 153.8 g/mol), and 4.00 grams of C S2 (molar mass = 76.1 g/mol). The mole fraction of benzene in this solution is:
A) 0.051
B) 0.200
C) 0.333
D) 0.395
E) 0.605
A) 0.051
B) 0.200
C) 0.333
D) 0.395
E) 0.605
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37
What is the mole fraction of lactose in a mixture prepared by adding 15.0 g of lactose (C12H22O11) to 100.0 g of water?
A) c lactose = 7.83×10 - 3
B) c lactose = 7.89×10 - 3
C) c lactose = 0.130
D) c lactose = 0.150
E) c lactose = 0.992
A) c lactose = 7.83×10 - 3
B) c lactose = 7.89×10 - 3
C) c lactose = 0.130
D) c lactose = 0.150
E) c lactose = 0.992
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38
Calculate the mole fraction of water in a solution made by mixing 1.0 g of water and 2.0 g of ethyl alcohol (C2H5OH).
A) 0.33
B) 0.25
C) 0.56
D) 0.78
E) 0.056
A) 0.33
B) 0.25
C) 0.56
D) 0.78
E) 0.056
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39
What is the mass percent , %(m/m), of an aqueous solution prepared by dissolving 12.5 g of KNO3 in 125 g of water?
A) 0.0909%
B) 0.100%
C) 9.09%
D) 10.0%
E) 90.9%
A) 0.0909%
B) 0.100%
C) 9.09%
D) 10.0%
E) 90.9%
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40
A solution is made by dissolving 0.750 moles of isopropanol (C3H7OH) in 0.250 moles of water. What is the mass percentage of isopropanol in this mixture?
MM(C3H7OH) = 60.10 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) 5.51% (m/m)
B) 9.1% (m/m)
C) 75.0% (m/m)
D) 76.9% (m/m)
E) 90.9% (m/m)
MM(C3H7OH) = 60.10 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) 5.51% (m/m)
B) 9.1% (m/m)
C) 75.0% (m/m)
D) 76.9% (m/m)
E) 90.9% (m/m)
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41
How many grams of NH3 are present in 2.50 L of 3.00 M NH3 solution?
MM(NH3) = 17.03 g/mol
A) 0.0705 g NH3
B) 7.50 g NH3
C) 14.2 g NH3
D) 20.4 g NH3
E) 128 g NH3
MM(NH3) = 17.03 g/mol
A) 0.0705 g NH3
B) 7.50 g NH3
C) 14.2 g NH3
D) 20.4 g NH3
E) 128 g NH3
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42
A solution of ethanol (C2H5OH) in water is 15.0% ethanol by weight. The molality of this solution is:
A) 15.0 m
B) 3.83 m
C) 3.25 m
D) 1.77 m
E) none of these
A) 15.0 m
B) 3.83 m
C) 3.25 m
D) 1.77 m
E) none of these
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43
What is the molality of a solution prepared by mixing 22 g of ethylene glycol (molar mass = 62 g/mol) with 108 g of water?
A) 0.17 m
B) 0.20 m
C) 3.3 m
D) 12. m
E) none of these
A) 0.17 m
B) 0.20 m
C) 3.3 m
D) 12. m
E) none of these
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44
How many moles of sucrose, C12H22O11, is present in 80.0 grams of an aqueous solution that is 3.50% sucrose by mass?
MM(C12H22O11) = 342.30 g/mol
A) 8.18×10 - 3 moles
B) 0.150 moles
C) 2.80 moles
D) 122 moles
E) 958 moles
MM(C12H22O11) = 342.30 g/mol
A) 8.18×10 - 3 moles
B) 0.150 moles
C) 2.80 moles
D) 122 moles
E) 958 moles
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45
What is the molality of a solution prepared by dissolving 76.5 grams of glycine (NH2CH2COOH) in 1.280 kg of water?
MM(glycine) = 75.07 g/mol
A) 0.752 m
B) 0.796 m
C) 1.02 m
D) 1.26 m
E) 59.8 m
MM(glycine) = 75.07 g/mol
A) 0.752 m
B) 0.796 m
C) 1.02 m
D) 1.26 m
E) 59.8 m
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46
The density of a solution that is 18.0% by mass lead nitrate [Pb(NO3)2] is 1.18 g/cm3. Calculate the molarity of the lead nitrate solution.
A) 4.9×10 - 7 M
B) 0.64 M
C) 12.0 M
D) 19.8 M
E) 15.3 M
A) 4.9×10 - 7 M
B) 0.64 M
C) 12.0 M
D) 19.8 M
E) 15.3 M
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47
What is the molality of a solution prepared by dissolving 10.5 grams of benzene, C6H6, in 18.5 grams of carbon tetrachloride, CCl4?
A) 0.568 m
B) 1.76 m
C) 4.64 m
D) 7.27 m
E) 11.5 m
A) 0.568 m
B) 1.76 m
C) 4.64 m
D) 7.27 m
E) 11.5 m
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48
How many grams of sulfur, S8 must be dissolved in 100. grams of naphthalene, C10H8, to make a 0.180 m solution?
Molar Mass(S8) = 256.48 g/mol
A) 0.0180 g S8
B) 4.62 g S8
C) 18.0 g S8
D) 142 g S8
E) 4330 g S8
Molar Mass(S8) = 256.48 g/mol
A) 0.0180 g S8
B) 4.62 g S8
C) 18.0 g S8
D) 142 g S8
E) 4330 g S8
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49
What is the molality of a solution prepared by mixing 11 g of C2H5OH (molar mass = 46 g/mol) witH₂ 04 g of water?
A) 1.9 m
B) 1.2 m
C) 3.3 m
D) 0.12 m
E) none of these
A) 1.9 m
B) 1.2 m
C) 3.3 m
D) 0.12 m
E) none of these
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50
How many moles of solute are present in 50.0 grams of solution that is 2.50% by mass glucose (C6H12O6)?
MM(C6H12O6) = 180.16 g/mol
A) 6.94×10 - 3 moles
B) 0.800 moles
C) 1.25 moles
D) 11.1 moles
E) 225 moles
MM(C6H12O6) = 180.16 g/mol
A) 6.94×10 - 3 moles
B) 0.800 moles
C) 1.25 moles
D) 11.1 moles
E) 225 moles
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51
A solution contains 1.00 mole of NaCl in 500 g of water. What is the molal concentration of NaCl?
A) 10.5 m
B) 2.00 m
C) 1.00 m
D) 0.0347 m
E) none of these
A) 10.5 m
B) 2.00 m
C) 1.00 m
D) 0.0347 m
E) none of these
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52
What is the molality of an aqueous 3.0% hydrogen peroxide solution?
(H2O2, molar mass = 34.0 g/mol)
A) 0.016 m
B) 0.030 m
C) 0.88 m
D) 0.91 m
E) none of these
(H2O2, molar mass = 34.0 g/mol)
A) 0.016 m
B) 0.030 m
C) 0.88 m
D) 0.91 m
E) none of these
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53
What mass of CsCl must be added to 0.55 liters of water to produce a 0.430 m solution?
Molar Mass(CsCl) = 168.35 g/mol and Density(H2O) = 1.00 g/mL
A) 0.237 g CsCl
B) 0.782 g CsCl
C) 39.8 g CsCl
D) 132 g CsCl
E) 215 g CsCl
Molar Mass(CsCl) = 168.35 g/mol and Density(H2O) = 1.00 g/mL
A) 0.237 g CsCl
B) 0.782 g CsCl
C) 39.8 g CsCl
D) 132 g CsCl
E) 215 g CsCl
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54
How many grams of LiNO3 must be added to 235 grams of water to prepare a 40.0% (m/m) solution of LiNO3?
A) 0.170 g LiNO3
B) 94.0 g LiNO3
C) 157 g LiNO3
D) 353 g LiNO3
E) 588 g LiNO3
A) 0.170 g LiNO3
B) 94.0 g LiNO3
C) 157 g LiNO3
D) 353 g LiNO3
E) 588 g LiNO3
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55
How many grams of water must be added tO5 0.0 g of glucose in order to prepare a 5.00% (m/m) glucose solution?
A) 2.5 g of H2O
B) 2.63 g of H2O
C) 250 g of H2O
D) 950 g of H2O
E) 1000 g of H2O
A) 2.5 g of H2O
B) 2.63 g of H2O
C) 250 g of H2O
D) 950 g of H2O
E) 1000 g of H2O
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56
How many grams of water must be added to 20.0 g of NaOH in order to prepare a 6.75% (m/m) solution?
A) 1.35 g H2O
B) 1.45 g H2O
C) 93.25 g H2O
D) 276 g H2O
E) 296 g H2O
A) 1.35 g H2O
B) 1.45 g H2O
C) 93.25 g H2O
D) 276 g H2O
E) 296 g H2O
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57
How many grams of glucose (C6H12O6) must be dissolved in 750 g of ethanol (C2H5OH) to prepare a 0.545 m solution?
MM(C6H12O6) = 180.15 g/mol
A) 2.27×10 - 3 g glucose
B) 73.6 g glucose
C) 131 g glucose
D) 248 g glucose
E) 441 g glucose
MM(C6H12O6) = 180.15 g/mol
A) 2.27×10 - 3 g glucose
B) 73.6 g glucose
C) 131 g glucose
D) 248 g glucose
E) 441 g glucose
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58
What is the molality of a solution that contains 9.00 g of ethylene glycol (C2H6O2) in 100 g of water?
A) 1.45 m
B) 1.61 m
C) 9.00 m
D) cannot be calculated
E) none of these
A) 1.45 m
B) 1.61 m
C) 9.00 m
D) cannot be calculated
E) none of these
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59
How many grams of NaCl must be dissolved in 500 g of H2O to prepare a 0.750 m solution?
Molar Mass(NaCl) = 58.44 g/mol
A) 6.42×10 - 3 g NaCl
B) 0.156 g NaCl
C) 21.9 g NaCl
D) 39.0 g NaCl
E) 87.7 g NaCl
Molar Mass(NaCl) = 58.44 g/mol
A) 6.42×10 - 3 g NaCl
B) 0.156 g NaCl
C) 21.9 g NaCl
D) 39.0 g NaCl
E) 87.7 g NaCl
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60
How many grams of glucose must be added to 275 g of water in order to prepare a 31.0% aqueous glucose solution?
A) 7.78 g glucose
B) 85.3 g glucose
C) 124 g glucose
D) 612 g glucose
E) 887 g glucose
A) 7.78 g glucose
B) 85.3 g glucose
C) 124 g glucose
D) 612 g glucose
E) 887 g glucose
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61
A research worker has only 300 g of a valuable solvent and wishes to use all of it to make up a 10.0 mass percent solution of solute A. How many grams of A should be weighed out?
A) 30.0 g
B) 22.2 g
C) 33.3 g
D) 10.0 g
E) cannot calculate without knowing molar masses
A) 30.0 g
B) 22.2 g
C) 33.3 g
D) 10.0 g
E) cannot calculate without knowing molar masses
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62
An aqueous solution of hydrochloric acid contains 36.0% (m/m). What is the molality of this same solution?
MM(HCl) = 36.46 g/mol and MM(H2O) = 18.015 g/mol
A) 0.217 m
B) 0.563 m
C) 9.87 m
D) 15.4 m
E) 98.7 m
MM(HCl) = 36.46 g/mol and MM(H2O) = 18.015 g/mol
A) 0.217 m
B) 0.563 m
C) 9.87 m
D) 15.4 m
E) 98.7 m
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63
Concentrated red fuming Nitric acid, HNO3, is 90.0% (m/m). What is this solution's Molar concentration if its density is 1.490 g/mL?
MM(HNO3) = 63.01 g/mol
A) 1.43 M
B) 2.36 M
C) 9.59 M
D) 14.3 M
E) 21.3 M
MM(HNO3) = 63.01 g/mol
A) 1.43 M
B) 2.36 M
C) 9.59 M
D) 14.3 M
E) 21.3 M
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64
An aqueous solution of concentrated glacial acetic acid, HC2H3O2, has a mole fraction of 0.993. What is this solution's Molar concentration if its density is 1.049 g/mL?
Molar Mass(HC2H3O2) = 60.05 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) 1.05 M
B) 6.31 M
C) 15.8 M
D) 16.6 M
E) 17.4 M
Molar Mass(HC2H3O2) = 60.05 g/mol and Molar Mass(H2O) = 18.02 g/mol
A) 1.05 M
B) 6.31 M
C) 15.8 M
D) 16.6 M
E) 17.4 M
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65
What is the mole fraction of NH3 for a 5.00% (m/m) aqueous solution of NH3?
A) c NH3 = 0.0527
B) c NH3 = 0.190
C) c NH3 = 0.294
D) c NH3 = 0.810
E) c NH3 = 0.947
A) c NH3 = 0.0527
B) c NH3 = 0.190
C) c NH3 = 0.294
D) c NH3 = 0.810
E) c NH3 = 0.947
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66
Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution that is 85% by mass ethylene glycol in water.
A) 1.4
B) 0.85
C) 0.18
D) 0.38
E) 0.62
A) 1.4
B) 0.85
C) 0.18
D) 0.38
E) 0.62
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67
Caffeine (C8H10N4O2) is a stimulant found in coffee and tea. What is the mass percent of caffeine, %(m/m), for a solution of caffeine dissolved in chloroform (CHCl3) that has a concentration of 0.0890 m ?
Molar Mass(C8H10N4O2) = 194.19 g/mol
A) 1.05%
B) 1.70%
C) 8.9%
D) 16.2%
E) 68.6%
Molar Mass(C8H10N4O2) = 194.19 g/mol
A) 1.05%
B) 1.70%
C) 8.9%
D) 16.2%
E) 68.6%
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68
A 2.90 m aqueous solution of acetic acid, C2H4O2 (60.0 g/mol), has a density of 1.020 g/mL. Calculate the molarity of the solution.
A) 2.90 M
B) 2.42 M
C) 2.96 M
D) 2.84 M
E) 2.52 M
A) 2.90 M
B) 2.42 M
C) 2.96 M
D) 2.84 M
E) 2.52 M
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69
Commercial concentrated ammonium hydroxide is 28.0% (m/m) NH3 and has a density of 0.890 g/mL. What is the molarity of a concentrated solution of ammonium hydroxide?
MW(NH3) = 17.03 g/mol
A) 5.4 M
B) 14.6 M
C) 16.4 M
D) 18.5 M
E) 22.8 M
MW(NH3) = 17.03 g/mol
A) 5.4 M
B) 14.6 M
C) 16.4 M
D) 18.5 M
E) 22.8 M
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70
A 2.20 molar solution of acetic acid (molar mass = 60.05 g/mol) has a density of 1.0178 g/mL. What is the molality of the solution?
A) 2.16 m
B) 2.20 m
C) 2.23 m
D) 2.48 m
E) none of these
A) 2.16 m
B) 2.20 m
C) 2.23 m
D) 2.48 m
E) none of these
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71
What is the molality of a 5.00% aqueous solution of NH3?
MM(NH3) = 17.03 g/mol
A) 1.12 m
B) 2.94 m
C) 3.09 m
D) 35.9 m
E) 52.6 m
MM(NH3) = 17.03 g/mol
A) 1.12 m
B) 2.94 m
C) 3.09 m
D) 35.9 m
E) 52.6 m
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72
Glacial acetic acid, HC2H3O2, is 99.8% (m/m). What is this solution's Molar concentration if its density is 1.049 g/mL?
MM(HC2H3O2) = 60.05 g/mol
A) 1.05 M
B) 6.31 M
C) 15.8 M
D) 16.6 M
E) 17.4 M
MM(HC2H3O2) = 60.05 g/mol
A) 1.05 M
B) 6.31 M
C) 15.8 M
D) 16.6 M
E) 17.4 M
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73
A 1.14 molar solution of ammonium chloride (molar mass = 53.50 g/mol) has a density of 1.019 g/mL. What is the molality of the solution?
A) 1.12 m
B) 1.16 m
C) 1.19 m
D) 1.24 m
E) none of these
A) 1.12 m
B) 1.16 m
C) 1.19 m
D) 1.24 m
E) none of these
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74
A concentrated reagent is sold as a solution that is 60.0 mass % solute. If the molar mass of the solute is 120 g/mol and the density of this solution is 1.70 g/mL, what is the molarity of the solution?
A) 18.0 M
B) 23.6 M
C) 32.5 M
D) 5.35 M
E) 8.50 M
A) 18.0 M
B) 23.6 M
C) 32.5 M
D) 5.35 M
E) 8.50 M
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75
The molality of an unknown aqueous solution was found to be 3.50 m . What is the mole fraction of water (molar mass = 18.0 g/mol) in this solution?
A) 0.059
B) 0.941
C) 0.778
D) 0.222
E) cannot be determined
A) 0.059
B) 0.941
C) 0.778
D) 0.222
E) cannot be determined
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76
What is the Molarity of a 5.00% (m/m) aqueous NH3 solution?
(Density of solution = 0.9651 g/mL)
A) 2.83 M
B) 2.94 M
C) 3.09 M
D) 3.20 M
E) 48.3 M
(Density of solution = 0.9651 g/mL)
A) 2.83 M
B) 2.94 M
C) 3.09 M
D) 3.20 M
E) 48.3 M
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77
A 3.60 m solution of KBr (molar mass = 119 g/mol) in water has a density of 1.26 g/mL. The molarity of this solution is:
A) 3.60 M
B) 2.86 M
C) 3.18 M
D) 4.09 M
E) 2.51 M
A) 3.60 M
B) 2.86 M
C) 3.18 M
D) 4.09 M
E) 2.51 M
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78
A 3.38 molar solution of cesium chloride (molar mass = 168.4 g/mol) has a density of 1.423 g/mL. What is the molality of the solution?
A) 4.81 m
B) 3.96 m
C) 3.38 m
D) 2.38 m
E) none of these
A) 4.81 m
B) 3.96 m
C) 3.38 m
D) 2.38 m
E) none of these
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79
What is the molality of a 10.0% aqueous solution of NH3?
MM(NH3) = 17.03 g/mol
A) 1.11 m
B) 1.70 m
C) 1.89 m
D) 5.87 m
E) 6.52 m
MM(NH3) = 17.03 g/mol
A) 1.11 m
B) 1.70 m
C) 1.89 m
D) 5.87 m
E) 6.52 m
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80
A solution of 3.0 g of citric acid (C6H8O7) dissolved in 70.0 g of water was found to have a density of 1.13 g/mL. The molarity of the solution was:
A) 1.81 M
B) 0.16 M
C) 11 M
D) 88 M
E) 0.24 M
A) 1.81 M
B) 0.16 M
C) 11 M
D) 88 M
E) 0.24 M
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