Question 1

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 2

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 3

A 17.2 g sample of potassium chlorate is dissolved in 250 g of water at 65 °C.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

When a solution is unsaturated, it means that it can dissolve more solute at the given temperature. In this case, no precipitate is observed after cooling the solution, which indicates that all of the potassium chlorate remained dissolved. Therefore, the solution is unsaturated.

Question 4

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

Accepted Answer

The answer of The mole fraction of urea (MW =...

Question 5

What is the mole fraction of LiCl in solution that is 9.0% by mass LiCl and has a density of 1.00 g/mL?
A)0.960
B)2.33
C)0.0403
D)2.12
E)9.00

Accepted Answer

To find the mole fraction of LiCl, we need to first find the moles of LiCl present in the solution. We can do this by assuming we have 100 g of the solution and then calculating the mass of LiCl present in it:

Mass of LiCl = 9.0 g (given as 9.0% by mass)
Mass of solvent (water) = 91.0 g (100 g - 9.0 g)

Now, we can convert the mass of LiCl to moles:

Moles of LiCl = 9.0 g / 42.39 g/mol (molar mass of LiCl) = 0.2125 mol

Next, we need to find the moles of the solvent (water). We can do this by dividing the mass of the solvent by its density:

Volume of solvent = Mass of solvent / Density = 91.0 mL

Moles of H2O = Volume of solvent / molar volume of water
= 91.0 mL / 18.015 g/mol
= 5.051 mol

Finally, we can find the mole fraction of LiCl:

Mole fraction of LiCl = Moles of LiCl / Total moles
= 0.2125 mol / (0.2125 mol + 5.051 mol)
= 0.0403

Thus, the mole fraction of LiCl in the given solution is 0.0403 or 4.03%. Therefore, the correct answer is C.

Question 6

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water.The density of the resulting solution is 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

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

Question 7

The molarity of urea (MW = 60.0 g/mol)in a solution prepared by dissolving 11 g of urea in of H₂O is ________ M.Assume the volume of the solution does not change when dissolving the urea. A)0.314 B)5.24 C)314 D)0.00524 E)0.00642

Accepted Answer

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

Question 8

A solution at that is 3.75 M in MnSO₄ monohydrate is considered a(n)________ solution.The solubility of MnSO₄ monohydrate in water at 20 °C is 70.0 g per 100.0 mL of water. A)hydrated B)solvated C)saturated D)unsaturated E)supersaturated

Accepted Answer

The answer of A solution at  that is 3.75...

Question 9

A solution is prepared by dissolving 24.7 g of KBr in 375 g water at 20.0 °C.What is the molarity (M)of KBr in this solution? Density of the solution at 20.0 °C is 0.998 g/mL.
A)0.552
B)0.0659
C)0.000552
D)552
E)65.9

Accepted Answer

To find the molarity (M) of KBr in the solution, we first need to calculate the number of moles of KBr and the volume of the solution in liters. The molar mass of KBr (K = 39.1 g/mol, Br = 79.9 g/mol) is approximately 119 g/mol. The number of moles of KBr is calculated as $ \frac{24.7 \, \text{g}}{119 \, \text{g/mol}} \approx 0.2075 \, \text{moles} $.The total mass of the solution is the mass of KBr plus the mass of water, which is $24.7 \, \text{g} + 375 \, \text{g} = 399.7 \, \text{g}$. Given the density of the solution is 0.998 g/mL, the volume of the solution can be calculated as $ \frac{399.7 \, \text{g}}{0.998 \, \text{g/mL}} \approx 400.5 \, \text{mL} $, or $0.4005 \, \text{L}$.Finally, the molarity (M) of KBr in the solution is $ \frac{0.2075 \, \text{moles}}{0.4005 \, \text{L}} \approx 0.518 \, \text{M} $. However, considering the choices provided and rounding, the closest answer is A) 0.552 M.

Question 10

A 30.0 g sample of potassium nitrate is dissolved in 100 g of water at 60 °C.The solution is cooled to 20.0 °C and a small amount of precipitate is observed.This solution is ________.
A)hydrated
B)placated
C)saturated
D)unsaturated
E)supersaturated

Accepted Answer

At 60 °C, the solubility of potassium nitrate in water is higher than at 20 °C. Therefore, when the solution is cooled, the solubility of potassium nitrate decreases and some of the excess potassium nitrate comes out of solution and forms a precipitate. At this point, the solution is saturated, meaning that it contains the maximum amount of dissolved solute at that temperature.

Question 11

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 12

What is the concentration (M)of HCl in a solution that is prepared by dissolving 25.5 g of HCl in of water? (Assume volume does not change after adding HCl.) A)1.28 B)0.0350 C)35.0 D)3.50 × 10^-5 E)14.0

Accepted Answer

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

25.5 g HCl x (1 mol HCl/36.46 g HCl) = 0.6998 mol HCl

Next, we calculate the total volume of the solution:

11.7 L water (given) + 0 L HCl = 11.7 L total solution

Now we can calculate the molarity (M) of the HCl solution:

M = mol/L 
M = 0.6998 mol / 11.7 L 
M = 0.0350 M

Therefore, the answer is B) 0.0350.

Question 13

A solution is prepared by dissolving 15.8 g of MgCl₂ in 255 g of water.What is the mole fraction of Cl^- in this solution? A)0.977 B)0.0116 C)0.0232 D)0.988 E)0.00583

Accepted Answer

First, we need to calculate the moles of MgCl2 in the solution:

moles MgCl2 = mass / molar mass = 15.8 g / 95.21 g/mol = 0.166 moles

Next, we need to calculate the mole fraction of Cl- in the solution. MgCl2 dissociates into Mg2+ and 2 Cl- ions in water.

moles Cl- = 2 x 0.166 moles = 0.332 moles

Total moles of solute and solvent = 0.166 moles + (255 g / 18.015 g/mol) = 14.0 moles

Mole fraction of Cl- = moles Cl- / total moles = 0.332 moles / 14.0 moles = 0.0237.

Rounded to three significant figures, the answer is 0.0232, which corresponds to choice C.

Question 14

The solubility of MnSO₄ monohydrate in water at 20 °C is 70.0 g per 100.0 mL of water.A solution at 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

A solution that is 0.401 M in MnSO₄ monohydrate has not reached the maximum amount of solute that can dissolve in the solvent (water) at that temperature (20 °C). Therefore, it is an unsaturated solution. The fact that it is a monohydrate means that it contains one molecule of water per formula unit of MnSO₄, but this does not affect the classification of the solution as unsaturated.

Question 15

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 16

What is the mole fraction of HCl in a solution that is prepared by dissolving 25.5 g of HCl in  of water? The density of the solution is 1.0 g/mL.
A)0.0155
B)0.0319
C)0.984
D)0.874
E)32.0

Accepted Answer

The answer of What is the mole fraction of HCl...

Question 17

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 the mass and density given: Volume = mass/density = 265.13 mL Next, we need to convert the mass of NH₃ to moles: Moles of NH₃ = 15.0 g / 328.24 g/mol = 0.0457 mol Finally, we can calculate the molarity of the solution: Molarity = moles of NH₃ / volume of solution in L Molarity = 0.0457 mol / (0.26513 L x 10^-3) = 3.24 M Therefore, the answer is D.

Question 18

A solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water.The density of the resulting solution is .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 amount of CaCl2 dissolved in water:

moles of CaCl2 = 23.7 g / 110.98 g/mol = 0.2134 mol

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

V = m / ρ = 375 g / 1.17 g/mL = 320.5 mL = 0.3205 L

Then, we can use the formula for molarity:

Molarity = moles of solute / liters of solution

Molarity of Cl- = 2 x Molarity of CaCl2 = 2 x 0.2134 mol / 0.3205 L = 1.12 M

Therefore, the answer is (C) 1.12.

Question 19

What is the molality of LiCl in solution that is 9.0% by mass LiCl and has a density of 1.00 g/mL?
A)9.00
B)2.12
C)2.33
D)90.0
E)0.0900

Accepted Answer

First, we need to convert the mass percent to molarity.

Assuming we have 100 g of solution: 
- 9.0 g is LiCl
- 91.0 g is water

We can calculate the moles of LiCl: 
moles LiCl = 9.0 g / 42.39 g/mol (molar mass of LiCl) = 0.2123 mol

Next, we need to calculate the mass of water (in kg) in the solution: 
mass water = 91.0 g = 0.091 kg

Now we can calculate the molality: 
molality = moles LiCl / mass of water (in kg) 
molality = 0.2123 mol / 0.091 kg = 2.33 mol/kg

Therefore, the answer is (C) 2.33.

Question 20

A solution is prepared by dissolving 25.0 g of NaCl in 500.0 g of water.What is the molality (m)of NaCl in the solution? The density of the solution is 1.0 g/mL.
A)0.000856
B)0.0500
C)50.0
D)8.56
E)0.856

Accepted Answer

Molality (m) is calculated as moles of solute per kilogram of solvent. First, calculate moles of NaCl: 25.0 g / 58.44 g/mol = 0.428 moles. Then, convert 500.0 g of water to kg: 500.0 g = 0.500 kg. Finally, molality = 0.428 moles / 0.500 kg = 0.856 m.

Quiz 13: Properties of Solutions

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 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 17.2 g sample of potassium chlorate is dissolved in 250 g of water at 65 °C.The solution is cooled to 30.0 °C and no precipitate is observed.This solution is ________.

The mole fraction of urea (MW = 60.0 g/mol) in a solution prepared by dissolving 16 g of urea in $The mole fraction of urea (MW = 60.0 g/mol) in a solution prepared by dissolving 16 g of urea in of H<sub>2</sub>O is . A) 0.58 B) 0.37 C) 0.13 D) 0.11 E) 9.1$ of H₂O is .

What is the mole fraction of LiCl in solution that is 9.0% by mass LiCl and has a density of 1.00 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 The concentration of CaCl₂ in this solution is ________ molal.

The molarity of urea (MW = 60.0 g/mol) in a solution prepared by dissolving 11 g of urea in of H₂O is ________ M.Assume the volume of the solution does not change when dissolving the urea.

A solution at that is 3.75 M in MnSO₄ monohydrate is considered a(n) ________ solution.The solubility of MnSO₄ monohydrate in water at 20 °C is 70.0 g per 100.0 mL of water.

A solution is prepared by dissolving 24.7 g of KBr in 375 g water at 20.0 °C.What is the molarity (M) of KBr in this solution? Density of the solution at 20.0 °C is 0.998 g/mL.

A 30.0 g sample of potassium nitrate is dissolved in 100 g of water at 60 °C.The solution is cooled to 20.0 °C and a small amount of 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.

What is the concentration (M) of HCl in a solution that is prepared by dissolving 25.5 g of HCl in of water? (Assume volume does not change after adding HCl.)

A solution is prepared by dissolving 15.8 g of MgCl₂ in 255 g of water.What is the mole fraction of Cl^- in this solution?

The solubility of MnSO₄ monohydrate in water at 20 °C is 70.0 g per 100.0 mL of water.A solution at 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 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 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 solution is prepared by dissolving 23.7 g of CaCl₂ in 375 g of water.The density of the resulting solution is .The concentration of Cl^- in this solution is ________ M.

What is the molality of LiCl in solution that is 9.0% by mass LiCl and has a density of 1.00 g/mL?

A solution is prepared by dissolving 25.0 g of NaCl in 500.0 g of water.What is the molality (m) of NaCl in the solution? The density of the solution is 1.0 g/mL.

Quiz 13: Properties of Solutions

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 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 17.2 g sample of potassium chlorate is dissolved in 250 g of water at 65 °C.The solution is cooled to 30.0 °C and no precipitate is observed.This solution is ________.

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

What is the mole fraction of LiCl in solution that is 9.0% by mass LiCl and has a density of 1.00 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 The concentration of CaCl2 in this solution is ________ molal.

The molarity of urea (MW = 60.0 g/mol) in a solution prepared by dissolving 11 g of urea in of H2O is ________ M.Assume the volume of the solution does not change when dissolving the urea.

A solution at that is 3.75 M in MnSO4 monohydrate is considered a(n) ________ solution.The solubility of MnSO4 monohydrate in water at 20 °C is 70.0 g per 100.0 mL of water.