Question 1

96.5 grams of an unknown substance were diluted with water in a 1.000 L volumetric flask. It was determined that the concentration of the substance was 1.22 M. What is the molar mass of the unknown substance?

A) 117 g·mol^-¹
B) 79.1 g·mol^-¹
C) 1.26´10^-²g·mol^-¹
D) 65.4 g·mol^-¹

79.1 g·mol^-¹

Accepted Answer

79.1 g·mol^-¹

Question 2

Equal volumes of ethanol (C₂H₅OH) and water are mixed. If the density of water is 1.00 g·cm^-³ and that of ethanol is 0.789, what is the mole fraction of ethanol in the mixture? A) 0.764 B) 0.309 C) 0.324 D) 0.236

Accepted Answer

0.236&#10;

Question 3

120.0 grams of calcium nitrate are weighed out and placed into a 1.5 L volumetric flask and the flask filled to the mark. What is the concentration of the nitrate ions in solution?&#10;A) 0.9751&#10;B) 0.4875&#10;C) 1.026&#10;D) 2.051

Accepted Answer

The correct answer is A because the concentration of the nitrate ions in solution can be calculated using the formula:Concentration = (Moles of solute) / (Volume of solution in liters)The moles of solute can be calculated using the formula:Moles of solute = (Mass of solute in grams) / (Molar mass of solute)The molar mass of calcium nitrate is 164.09 g/mol. Therefore, the moles of calcium nitrate in 120.0 grams is:Moles of calcium nitrate = 120.0 g / 164.09 g/mol = 0.731 molThe volume of the solution is 1.5 L. Therefore, the concentration of the nitrate ions in solution is:Concentration = 0.731 mol / 1.5 L = 0.4875 MThe concentration of the nitrate ions in solution is 0.4875 M, which is choice A.

Question 4

How many ml of 0.1506 M NaOH are required to reach the endpoint in a titration with 30.00 ml of 0.2215 M HCl?&#10;A) 20.40&#10;B) 30.15&#10;C) 31.18&#10;D) 44.12

Accepted Answer

The answer of How many ml of 0.1506 M NaOH...

Question 5

Which, if any, of the following would you expect to readily dissolve in water?&#10;A)&#10; &#10;B)&#10; &#10;C)  &#10;D) a and c&#10;E) none of the above

Accepted Answer

The answer of Which, if any, of the following would...

Question 6

What is the expected boiling point of a solution made by combining 400.0 g of water with 100.0 g of Na₂CO₃? A) 103.6°C B) 113.5°C C) 101.2°C D) 104.8°C

Accepted Answer

The boiling point elevation formula is ΔTb = iKbm, where ΔTb is the boiling point elevation, i is the van't Hoff factor (number of particles the solute breaks into), Kb is the ebullioscopic constant of the solvent (water in this case, which is 0.512°C·kg/mol), and m is the molality of the solution. Na₂CO is sodium bicarbonate (NaHCO3), which dissociates into 2 particles (Na+ and HCO3-), so i = 2. The molar mass of NaHCO3 is approximately 84 g/mol. The molality (m) = moles of solute/kg of solvent = (100 g / 84 g/mol) / (400 g / 1000 g/kg) = 1.19 mol/kg. Thus, ΔTb = 2 * 0.512°C·kg/mol * 1.19 mol/kg = 1.22°C. The normal boiling point of water is 100°C, so the expected boiling point of the solution is 100°C + 1.22°C = 101.22°C, which rounds to 101.2°C, matching choice C. However, my initial selection was incorrect; the correct choice is C, not A.

Question 7

The following question(s) pertain to the dissolution of 200.0 g of KMnO4 into 600.0 g. of water.

-Refer to Exhibit 11-2. What is the molality of the KMnO₄ in the 800.0 gram solution?

A) 1.255 m
B) 2.109 m
C) 1.054 m
D) 2.223 m

Accepted Answer

Molality (m) is calculated as moles of solute per kilogram of solvent. First, calculate the moles of KMnO4: 200.0 g / 158.04 g/mol = 1.265 moles. The mass of the solvent (water) is 600.0 g or 0.600 kg. Molality = 1.265 moles / 0.600 kg = 2.109 m.

Question 8

The following question(s) pertain to the dissolution of 200.0 g of KMnO4 into 600.0 g. of water.

-Refer to Exhibit 11-2. What is the mole fraction of MnO₄^- ions in the solution?

A) 0.0353
B) 0.0365
C) 0.0433
D) 0.0333

Accepted Answer

The mole fraction of MnO₄^- ions in the solution is 0.0353. This is because the mole fraction of a component in a solution is equal to the number of moles of that component divided by the total number of moles of all components in the solution. In this case, the number of moles of MnO₄^- ions is equal to the number of moles of KMnO₄ that was dissolved, which is 200.0 g / 158.034 g/mol = 1.265 mol. The total number of moles of all components in the solution is equal to the number of moles of KMnO₄ plus the number of moles of water, which is 600.0 g / 18.015 g/mol = 33.31 mol. Therefore, the mole fraction of MnO₄^- ions in the solution is 1.265 mol / (1.265 mol + 33.31 mol) = 0.0353.

Question 9

pertain to the dissolution of 200.0 g of KMnO4 into 600.0 g. of water.

- Refer to Exhibit 11-2. If 200.0 ml of a solution that was 0.20 M in KMnO4 were added to the 800.0 g solution, would you expect the boiling point of that solution to go up?

A) Yes
B) No
C) More data is needed to answer the question

Accepted Answer

The answer of pertain to the dissolution of 200.0 g...

Question 10

To prepare 0.250 L of a 0.115_M NaBr solution, a technician must mix A) 0.115 moles of NaBr with 0.250 L of H₂O. B) 0.115 moles of NaBr with 0.250 kg of H₂O. C) 0.115 moles of NaBr with enough H₂O to produce 0.250 kg of solution. D) 0.115 moles of NaBr with enough H₂O to produce 0.250 L of solution. E) none of these

Accepted Answer

The answer of To prepare 0.250 L of a 0.115_M...

Question 11

320 mL of 1.00 M HCl(aq) is combined with 1.18 L of 0.125 M HCl(aq) to give a total volume of 1.50 L. The molarity of the HCl in the final solution is&#10;A) 0.562 M&#10;B) 0.468 M&#10;C) 0.312 M&#10;D) 0.175 M&#10;E) none of these

Accepted Answer

The answer of 320 mL of 1.00 M HCl(aq) is...

Question 12

Acetone (mw = 58.08, P 25&#176;C =232 mmHg) and butanone (mw = 72.11, P 25 &#176;C =100 mmHg) have the indicated molar masses and vapor pressures. A container holds 1.00 kg of butanone. How much acetone must be added to the butanone to elevate the total vapor pressure over the mixture to 125 mmHg at 25&#176;C?&#10;A) 5.313 kg&#10;B) 290 g&#10;C) 188 g&#10;D) More information needed&#10;E) None of the above

Accepted Answer

The answer of Acetone (mw = 58.08, P 25&#176;C =232...

Deck 6: Solutions