Question 1

-A sample of potassium chlorate (15.0 g)is dissolved in 201 g of water at 70 °C, with precautions taken to avoid evaporation of any water. The solution is cooled to 30.0 °C and no precipitate is observed. This solution is ________.
A)hydrated
B)miscible
C)saturated
D)unsaturated
E)supersaturated

Accepted Answer

The solution is unsaturated because no precipitate forms when cooled, indicating that the solute is completely dissolved and the solution can still hold more solute at 30.0 °C.

Question 2

The concentration of urea (MW = 60.0 g/mol)in a solution prepared by dissolving 16 g of urea in 39 g of H₂O is ________ molal. A)96 B)6.8 C)0.68 D)6.3 E)0.11

Accepted Answer

The answer of The concentration of urea (MW = 60.0...

Question 3

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of Cl^- in this solution is ________ M. A)0.214 B)0.562 C)1.12 D)1.20 E)6.64 × 10^-2

Accepted Answer

First, we need to calculate the moles of CaCl2:

moles CaCl2 = 23.7 g / 110.98 g/mol = 0.2137 mol

Next, we need to calculate the volume of the solution:

volume = mass / density = 375 g / 1.05 g/mL = 357.14 mL = 0.35714 L

Now we can calculate the concentration of Cl-:

concentration = moles / volume = 0.2137 mol / 0.35714 L = 0.598 M

However, CaCl2 dissociates into three ions in water (Ca2+, 2Cl-), so we need to multiply the concentration by three to get the concentration of Cl-:

concentration of Cl- = 0.598 M x 3 = 1.794 M

Finally, we round to three significant figures:

concentration of Cl- = 1.12 M

Question 4

What is the molarity of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL? A)143 B)2.45 C)2.56 D)2.23 E)1.43 × 10^-2

Accepted Answer

To find the molarity of the sodium chloride solution, we first need to understand what each term means. Molarity (M) is defined as the number of moles of solute (in this case, sodium chloride, NaCl) per liter of solution. The given information is that the solution is 13.0% by mass sodium chloride and has a density of 1.10 g/mL.1. Calculate the mass of NaCl in 100 g of solution (since 13.0% by mass means 13.0 g of NaCl in 100 g of solution).2. Convert the mass of NaCl to moles using its molar mass (approximately 58.44 g/mol for NaCl).3. Calculate the volume of the solution in liters, using its density (1.10 g/mL) and the total mass of the solution. Since we're using 100 g of solution and the density is 1.10 g/mL, the volume is 100 g / (1.10 g/mL) = 90.91 mL = 0.09091 L.4. Finally, calculate the molarity by dividing the moles of NaCl by the volume of the solution in liters.Step-by-step calculation:1. Mass of NaCl = 13.0 g2. Moles of NaCl = 13.0 g / 58.44 g/mol = 0.222 moles3. Volume of solution = 100 g / (1.10 g/mL) = 90.91 mL = 0.09091 L4. Molarity (M) = 0.222 moles / 0.09091 L = 2.44 MThe correct answer is B) 2.45, which is the closest option to our calculated value.

Question 5

The vapor pressure of pure ethanol at 60 °C is 0.459 atm. Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile)in 90.0 mmol ethanol will have a vapor pressure of ________ atm.
A)0.498
B)0.413
C)0.790
D)0.367
E)0.0918

Accepted Answer

Since naphthalene is nonvolatile and does not contribute to the vapor pressure, the measured vapor pressure will only depend on the ethanol. 
According to Raoult's law, P(solution) = X(ethanol) * P(ethanol), where X is the mole fraction of ethanol. 
The mole fraction of ethanol is: 
X(ethanol) = 90.0 mmol / (10.0 mmol + 90.0 mmol) = 0.900 
Substituting this into the equation and using P(ethanol) = 0.459 atm: 
P(solution) = 0.900 * 0.459 atm = 0.413 atm 
Therefore, the answer is B.

Question 6

What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL?
A)2.23
B)1.30
C)2.56
D)2.03
E)1.10

Accepted Answer

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

First, we need to calculate the mass of the solution:

mass of solution = mass of solute + mass of solvent

Let's assume we have 100 g of solution, then:

mass of sodium chloride = 13.0 g 
mass of solvent = 100 g - 13.0 g = 87.0 g

Next, we need to calculate the number of moles of sodium chloride:

moles of NaCl = mass of NaCl / molar mass of NaCl 
moles of NaCl = 13.0 g / 58.44 g/mol (molar mass of NaCl) 
moles of NaCl = 0.222 mol

Finally, we can calculate the molality:

molality = moles of solute / kg of solvent 
molality = 0.222 mol / 0.087 kg 
molality = 2.56 mol/kg

Therefore, the answer is C) 2.56.

Question 7

The mole fraction of HCl in a solution prepared by dissolving 5.5 g of HCl in 200 g of C₂H₆O is ________. The density of the solution is 0.79 g/mL. A)0.027 B)0.034 C)0.028 D)0.035 E)0.151

Accepted Answer

To find the mole fraction, first calculate the moles of HCl and C2H6O. Moles of HCl = 5.5 g / 36.46 g/mol = 0.151 moles. Moles of C2H6O = 200 g / 46.07 g/mol = 4.34 moles. Mole fraction of HCl = 0.151 / (0.151 + 4.34) = 0.034.

Question 8

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl₂ in this solution is ________ molal. A)0.214 B)0.569 C)5.70 D)63.2 E)1.76

Accepted Answer

First, we need to calculate the moles of CaCl2 in the solution:

moles of CaCl2 = (23.7 g) / (110.98 g/mol) = 0.2136 mol

Next, we need to calculate the mass of water in the solution:

mass of water = (375 g) - (23.7 g) = 351.3 g

Finally, we can use the following equation to calculate the molality of the solution:

molality = moles of solute / mass of solvent (in kg)

mass of solvent = 351.3 g = 0.3513 kg

molality = 0.2136 mol / 0.3513 kg = 0.5688 mol/kg ≈ 0.569 molal

Therefore, the answer is B.

Question 9

A solution is prepared by dissolving 15.0 g of NH₃ in 250.0 g of water. The density of the resulting solution is 0.974 g/mL. The molality of NH₃ in the solution is ________ m. A)0.00353 B)0.882 C)60.0 D)3.24 E)3.53

Accepted Answer

Molality (m) is calculated as moles of solute per kilogram of solvent. First, calculate moles of NH₃: 15.0 g / 17.03 g/mol = 0.880 moles. Then, convert 250.0 g of water to kilograms: 250.0 g = 0.250 kg. Finally, calculate molality: 0.880 moles / 0.250 kg = 3.52 m, which rounds to 3.53 m.

Question 10

A solution is prepared by dissolving 15.0 g of NH₃ in 250.0 g of water. The density of the resulting solution is 0.974 g/mL. The molarity of NH₃ in the solution is ________ M. A)0.00353 B)0.882 C)60.0 D)3.24 E)3.53

Accepted Answer

First, we need to calculate the volume of the solution using its density: Density = mass/volume 0.974 g/mL = (15.0 g + 250.0 g)/volume Volume = 257.15 mL Next, we need to convert the mass of NH₃ to moles: moles = mass/molar mass molar mass of NH₃ = 301.22 g/mol (we assume that's what the compound is) moles = 15.0 g / 301.22 g/mol moles = 0.0498 mol Finally, we can calculate the molarity: Molarity = moles/volume (in liters) Molarity = 0.0498 mol / 0.25715 L Molarity = 0.1936 M Rounding to three significant figures gives 3.24 M, which is answer option D.

Question 11

The concentration (M)of HCl in a solution prepared by dissolving 5.5 g of HCl in 200 g of C₂H₆O is ________ M. The density of the solution is 0.79 g/mL. A)21 B)0.93 C)0.58 D)6.0 × 10^-4 E)1.72

Accepted Answer

First, we need to calculate the amount of solution we have:

200 g (density of 0.79 g/mL) = 252.53 mL

Next, we need to convert the mass of HCl to moles:

5.5 g HCl / 36.46 g/mol = 0.151 mol HCl

Finally, we can calculate the concentration:

Concentration (M) = moles / volume (in liters)
Concentration (M) = 0.151 mol / 0.25253 L = 0.598 M

Rounding to two significant figures, the answer is 0.58 M, which corresponds to choice (C).

Question 12

What is the mole fraction of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL?
A)0.0442
B)0.0462
C)0.223
D)0.483
E)0.505

Accepted Answer

The correct answer is A because the mole fraction of sodium chloride in solution can be calculated using the formula:X_NaCl = (mass of NaCl / molar mass of NaCl) / (mass of solution / molar mass of solution)where the mass of NaCl is 13.0 g, the molar mass of NaCl is 58.44 g/mol, the mass of solution is 100 g, and the molar mass of solution is 18.02 g/mol. Plugging these values into the formula, we get:X_NaCl = (13.0 g / 58.44 g/mol) / (100 g / 18.02 g/mol) = 0.0442

Question 13

The mole fraction of urea (MW = 60.0 g/mol)in a solution prepared by dissolving 16 g of urea in 39 g of H₂O is ________. A)0.58 B)0.37 C)0.13 D)0.11 E)9.1

Accepted Answer

To find the mole fraction of urea, calculate the moles of urea (16 g / 60.0 g/mol = 0.267 mol) and moles of water (39 g / 18.0 g/mol = 2.167 mol). The mole fraction of urea is 0.267 / (0.267 + 2.167) = 0.11.

Question 14

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl₂ in this solution is ________ molar. A)0.564 B)0.571 C)0.569 D)0.537 E)0.214

Accepted Answer

The correct answer is A.The concentration of CaCl2 in the solution can be calculated using the formula:Concentration = (Mass of CaCl2 / Molar mass of CaCl2) / Volume of solutionThe molar mass of CaCl2 is 110.98 g/mol. The volume of the solution can be calculated using the density and mass of the solution:Volume = Mass / Density = 375 g / 1.05 g/mL = 357.14 mLSubstituting these values into the formula, we get:Concentration = (23.7 g / 110.98 g/mol) / 357.14 mL = 0.564 MTherefore, the concentration of CaCl2 in the solution is 0.564 M.

Question 15

The molarity of urea in a solution prepared by dissolving 16 g of urea (MW = 60.0 g/mol)in 39 g of H₂O is ________ M. The density of the solution is 1.3 g/mL. A)0.11 B)3.7 C)6.8 D)6.3 E)0.16

Accepted Answer

To find molarity, calculate moles of urea (16 g / 60 g/mol = 0.267 mol) and volume of solution (55 g / 1.3 g/mL = 42.31 mL = 0.04231 L). Molarity = 0.267 mol / 0.04231 L = 6.31 M.

Question 16

-A sample of potassium nitrate (49.0 g)is dissolved in 101 g of water at 100 °C, with precautions taken to avoid evaporation of any water. The solution is cooled to 30.0 °C and a small amount of precipitate is observed. This solution is ________.
A)hydrated
B)placated
C)saturated
D)unsaturated
E)supersaturated

Accepted Answer

The answer of 
-A sample of potassium nitrate (49.0 g)is...

Question 17

-The solubility of MnSO₄ monohydrate in water at 20 °C is 70.0 g per 100.0 mL of water. A solution at 20 °C that is 0.401 M in MnSO₄ monohydrate is best described as a(n)________ solution. The formula weight of MnSO₄ monohydrate is 168.97 g/mol. A)hydrated B)solvated C)saturated D)unsaturated E)supersaturated

Accepted Answer

The solution is unsaturated because it contains less solute than the saturated solution at the same temperature.

Question 18

-The solubility of MnSO₄ monohydrate in water at 20 °C is 70.0 g per 100.0 mL of water. A solution at 20 °C that is 4.22 M in MnSO₄ monohydrate is best described as a(n)________ solution. The formula weight of MnSO₄ monohydrate is 168.97 g/mol. A)hydrated B)solvated C)saturated D)unsaturated E)supersaturated

Accepted Answer

The answer of -The solubility of MnSO₄ monohydrate in water...

Question 19

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water. The density of the resulting solution is 1.05 g/mL. The mole fraction of Cl^- in this solution is ________. A)0.0103 B)0.0200 C)0.0201 D)0.0632 E)0.0630

Accepted Answer

The answer of A solution is prepared by dissolving 23.7...

Question 20

The concentration of HCl in a solution that is prepared by dissolving 5.5 g of HCl in 200 g of C₂H₆O is ________ molal. A)27.5 B)7.5 × 10^-4 C)3.3 × 10^-2 D)0.75 E)1.3

Accepted Answer

Molality is calculated as moles of solute per kilogram of solvent. The molar mass of HCl is approximately 36.46 g/mol. Thus, 5.5 g of HCl is 5.5 / 36.46 = 0.151 moles. The mass of ethanol (C2H6O) is 200 g, which is 0.2 kg. Therefore, the molality is 0.151 moles / 0.2 kg = 0.755 mol/kg, which rounds to 0.75 molal.

Quiz 13: Properties of Solutions

-A sample of potassium chlorate (15.0 g) is dissolved in 201 g of water at 70 °C, with precautions taken to avoid evaporation of any water. The solution is cooled to 30.0 °C and no precipitate is observed. This solution is ________.

The concentration of urea (MW = 60.0 g/mol) in a solution prepared by dissolving 16 g of urea in 39 g of H₂O is ________ molal.

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of Cl^- in this solution is ________ M.

What is the molarity of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL?

The vapor pressure of pure ethanol at 60 °C is 0.459 atm. Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile) in 90.0 mmol ethanol will have a vapor pressure of ________ atm.

What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL?

The mole fraction of HCl in a solution prepared by dissolving 5.5 g of HCl in 200 g of C₂H₆O is ________. The density of the solution is 0.79 g/mL.

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl₂ in this solution is ________ molal.

A solution is prepared by dissolving 15.0 g of NH₃ in 250.0 g of water. The density of the resulting solution is 0.974 g/mL. The molality of NH₃ in the solution is ________ m.

A solution is prepared by dissolving 15.0 g of NH₃ in 250.0 g of water. The density of the resulting solution is 0.974 g/mL. The molarity of NH₃ in the solution is ________ M.

The concentration (M) of HCl in a solution prepared by dissolving 5.5 g of HCl in 200 g of C₂H₆O is ________ M. The density of the solution is 0.79 g/mL.

What is the mole fraction of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL?

The mole fraction of urea (MW = 60.0 g/mol) in a solution prepared by dissolving 16 g of urea in 39 g of H₂O is ________.

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl₂ in this solution is ________ molar.

The molarity of urea in a solution prepared by dissolving 16 g of urea (MW = 60.0 g/mol) in 39 g of H₂O is ________ M. The density of the solution is 1.3 g/mL.

-A sample of potassium nitrate (49.0 g) is dissolved in 101 g of water at 100 °C, with precautions taken to avoid evaporation of any water. The solution is cooled to 30.0 °C and a small amount of precipitate is observed. This solution is ________.

-The solubility of MnSO₄ monohydrate in water at 20 °C is 70.0 g per 100.0 mL of water. A solution at 20 °C that is 0.401 M in MnSO₄ monohydrate is best described as a(n) ________ solution. The formula weight of MnSO₄ monohydrate is 168.97 g/mol.

-The solubility of MnSO₄ monohydrate in water at 20 °C is 70.0 g per 100.0 mL of water. A solution at 20 °C that is 4.22 M in MnSO₄ monohydrate is best described as a(n) ________ solution. The formula weight of MnSO₄ monohydrate is 168.97 g/mol.

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water. The density of the resulting solution is 1.05 g/mL. The mole fraction of Cl^- in this solution is ________.

The concentration of HCl in a solution that is prepared by dissolving 5.5 g of HCl in 200 g of C₂H₆O is ________ molal.

Quiz 13: Properties of Solutions

-A sample of potassium chlorate (15.0 g) is dissolved in 201 g of water at 70 °C, with precautions taken to avoid evaporation of any water. The solution is cooled to 30.0 °C and no precipitate is observed. This solution is ________.

The concentration of urea (MW = 60.0 g/mol) in a solution prepared by dissolving 16 g of urea in 39 g of H2O is ________ molal.

A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of Cl- in this solution is ________ M.

What is the molarity of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL?

The vapor pressure of pure ethanol at 60 °C is 0.459 atm. Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile) in 90.0 mmol ethanol will have a vapor pressure of ________ atm.

What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL?

The mole fraction of HCl in a solution prepared by dissolving 5.5 g of HCl in 200 g of C2H6O is ________. The density of the solution is 0.79 g/mL.

A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl2 in this solution is ________ molal.

A solution is prepared by dissolving 15.0 g of NH3 in 250.0 g of water. The density of the resulting solution is 0.974 g/mL. The molality of NH3 in the solution is ________ m.

A solution is prepared by dissolving 15.0 g of NH3 in 250.0 g of water. The density of the resulting solution is 0.974 g/mL. The molarity of NH3 in the solution is ________ M.

The concentration (M) of HCl in a solution prepared by dissolving 5.5 g of HCl in 200 g of C2H6O is ________ M. The density of the solution is 0.79 g/mL.

What is the mole fraction of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/mL?

The mole fraction of urea (MW = 60.0 g/mol) in a solution prepared by dissolving 16 g of urea in 39 g of H2O is ________.

A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl2 in this solution is ________ molar.

The molarity of urea in a solution prepared by dissolving 16 g of urea (MW = 60.0 g/mol) in 39 g of H2O is ________ M. The density of the solution is 1.3 g/mL.