Question 1

What mass of LiOH is required to prepare 0.250 L of a 3.55 M solution?
A) 0.0371 g
B) 0.888 g
C) 21.3 g
D) 250. g
E) 340. g

Accepted Answer

First, we need to use the formula:

Molarity (M) = moles of solute / liters of solution

We can rearrange this formula to solve for the moles of solute:

moles of solute = Molarity (M) x liters of solution

Now, we can calculate the moles of LiOH needed:

moles of LiOH = 3.55 mol/L x 0.250 L = 0.8875 mol

Finally, we can convert moles to grams using the molar mass of LiOH:

mass of LiOH = moles of LiOH x molar mass of LiOH

mass of LiOH = 0.8875 mol x 23.95 g/mol = 21.25 g

Therefore, the best choice is C, 21.3 g.

Question 2

What volume of concentrated (14.7 M) phosphoric acid is needed to prepare 25.0 L of 3.0 M H₃PO₄? A) 0.20 L B) 0.57 L C) 1.8 L D) 3.6 L E) 5.1 L

Accepted Answer

The answer of What volume of concentrated (14.7 M) phosphoric...

Question 3

Calcium nitrite is used as a corrosion inhibitor in lubricants. What is the molality of a solution prepared by dissolving 18.5 g of calcium nitrite in 83.5 g of distilled water?
A) 0.181 m
B) 0.222 m
C) 0.596 m
D) 1.35 m
E) 1.68 m

Accepted Answer

The answer of Calcium nitrite is used as a corrosion...

Question 4

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the aqueous solubility of atmospheric O₂ is 2.67 × 10^-4 M at sea level and 25°C, what is the solubility of O₂ at an elevation of 12,000 ft where the total pressure is 0.657 atm? Assume the temperature is 25°C, and that the mole fraction of O₂ in air is 0.209 at both 12,000 ft and at sea level. A) 1.75 × 10^-4 M B) 2.67 × 10^-4 M C) 3.66 × 10^-5 M D) 4.06 × 10^-4 M E) 5.58 × 10^-5 M

Accepted Answer

According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The solubility of oxygen at 12,000 ft can be calculated using the following equation:

Solubility at 12,000 ft = Solubility at sea level x (Partial pressure of O2 at 12,000 ft / Partial pressure of O2 at sea level)

The partial pressure of O2 at 12,000 ft can be calculated using the total pressure and the mole fraction of O2 in air:

Partial pressure of O2 at 12,000 ft = Total pressure at 12,000 ft x Mole fraction of O2 at 12,000 ft
                                     = 0.657 atm x 0.209
                                     = 0.137 atm

Similarly, the partial pressure of O2 at sea level can be calculated as:

Partial pressure of O2 at sea level = Total pressure at sea level x Mole fraction of O2 at sea level
                                      = 1 atm x 0.209
                                      = 0.209 atm

Substituting these values in the above equation:

Solubility at 12,000 ft = 2.67 × 10^-4 M x (0.137 atm / 0.209 atm) = 1.75 × 10^-4 M

Therefore, the solubility of oxygen at an elevation of 12,000 ft is 1.75 × 10^-4 M, which is closest to choice A.

Question 5

The solubility of nitrogen gas in water at 25°C and a nitrogen pressure of 522 mmHg is 4.7 × 10^-4 mol/L. What is the value of the Henry's law constant in mol L^-1• atm^-1? A) 6.8 × 10^-4 mol L^-1• atm^-1 B) 4.7 × 10^-4 mol L^-1• atm^-1 C) 3.2 × 10^-4 mol L^-1• atm^-1 D) 9.0 × 10^-7 mol L^-1• atm^-1 E) 1.5 × 10³ mol L^-1• atm^-1

Accepted Answer

Henry's law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The constant of proportionality is known as Henry's law constant (kH). The formula for Henry's law is S = kH * P, where S is the solubility of the gas, kH is Henry's law constant, and P is the pressure of the gas.Given:- Solubility (S) = 4.7 × 10^-4 mol/L- Pressure (P) = 522 mmHg = 522/760 atm (since 1 atm = 760 mmHg)To find kH, rearrange the formula to kH = S / P.kH = (4.7 × 10^-4 mol/L) / (522/760 atm) = (4.7 × 10^-4 mol/L) / (0.6868 atm) ≈ 6.8 × 10^-4 mol/L•atmTherefore, the Henry's law constant is 6.8 × 10^-4 mol/L•atm.

Question 6

Cadmium bromide is used in photography and lithography. What is the molality of a solution prepared by dissolving 45.38 g of CdBr₂ in 375.0 g of water? A) 0.03035 m B) 0.01600 m C) 0.1210 m D) 0.4446 m E) 16.00 m

Accepted Answer

The answer of Cadmium bromide is used in photography and...

Question 7

Copper(II) bromide is used as a wood preservative. What mass of CuBr₂ is needed to prepare 750.0 mL of a 1.25 M solution? A) 134 g B) 209 g C) 372 g D) 750 g E) 938 g

Accepted Answer

To find the mass of CuBr2 needed, use the formula: mass = molarity × volume × molar mass. The molar mass of CuBr2 is approximately 223.55 g/mol. For a 1.25 M solution in 750.0 mL (0.750 L), the mass is 1.25 mol/L × 0.750 L × 223.55 g/mol = 209 g.

Question 8

What is the molarity of a solution of 10.0% by mass cadmium sulfate, CdSO₄ (208.47 g/mol)? The density of the solution is 1.10 g/mL. A) 0.528 M B) 0.436 M C) 0.479 M D) 0.100 M E) 1.89 M

Accepted Answer

To find the molarity of the solution, we first determine the mass of cadmium sulfate in 1000 mL (or 1 L) of solution, since molarity is moles of solute per liter of solution. With a 10.0% by mass solution and a density of 1.10 g/mL, the mass of the solution per liter is 1100 g. Therefore, the mass of cadmium sulfate in this volume is 10.0% of 1100 g, which is 110 g. The molar mass of cadmium sulfate (CdSO4) is approximately 208.47 g/mol, so the number of moles of cadmium sulfate in 110 g is 110 g / 208.47 g/mol = 0.528 mol. Thus, the molarity of the solution is 0.528 M.

Question 9

The density of a 20.3 M CH₃OH (methanol) solution is 0.858 g/mL. What is the molality of this solution? H₂O is the solvent. A) 17.4 m B) 20.8 m C) 23.7 m D) 70.0 m E) 97.8 m

Accepted Answer

The answer of The density of a 20.3 M CH₃OH...

Question 10

Which law describes the quantitative relationship between the solubility of a gas and its partial pressure?
A) Entropy
B) Henry's law
C) Enthalpy
D) Boyle's law
E) Avogadro's law

Accepted Answer

Henry's law describes the quantitative relationship between the solubility of a gas and its partial pressure.

Question 11

At 10°C one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. What is the Henry's law constant of Cl₂ in water? (R = 0.0821 atm • L • mol^-1•K^-1 ) A) 3.8 mol • L^-1• atm^-1 B) 0.043 mol • L^-1• atm^-1 C) 36 mol • L^-1• atm^-1 D) 3.1 mol • L^-1• atm^-1 E) 0.13 mol • L^-1• atm^-1

Accepted Answer

The answer of At 10°C one volume of water dissolves...

Question 12

Which combination of a 36.0% (w/w) stock solution of acetic acid (d = 1.045 g/mL) and water will result in 1.00 kg of a 15.0% (w/w) acetic acid solution?
A) 417 mL of 36% acetic acid in 583 mL of distilled water
B) 417 g of 36% acetic acid in 583 g of distilled water
C) 360 mL of 36% acetic acid in 640 mL of distilled water
D) 360 g of 36% acetic acid in 640 g of distilled water
E) 150 g of 36% acetic acid in 850 g of distilled water

Accepted Answer

To achieve a 15.0% (w/w) acetic acid solution from a 36.0% solution, the mass of the acetic acid in the final solution must be 15% of the total mass (1.00 kg). Option B correctly maintains the mass proportions, using 417 g of the 36% acetic acid solution to achieve the desired concentration when mixed with 583 g of water, totaling 1.00 kg of the final solution.

Question 13

What is the molarity of a 17.0% by mass solution of sodium acetate, NaC₂H₃O₂ (82.0 g/mol), in water? The density of the solution is 1.09 g/mL. A) 0.170 M B) 0.207 M C) 2.07 M D) 2.26 M E) 1.90 M

Accepted Answer

To find the molarity of the solution, we first need to find the moles of sodium acetate present in 1 L of the solution.

Mass of NaC2H3O2 in 1 L of solution = (17.0% by mass) x (1.09 g/mL) x (1000 mL/L) = 186.53 g
Number of moles of NaC2H3O2 = (mass/molar mass) = 186.53 g / 82.0 g/mol = 2.275 mol

Therefore, the molarity of the solution is:

Molarity = number of moles / volume in liters = 2.275 mol/L

Answer: D) 2.26 M

Question 14

Calculate the molality of a 20.0% by mass ammonium sulfate (NH₄)₂SO₄ solution. The density of the solution is 1.117 g/mL. A) 0.151 m B) 0.200 m C) 0.223 m D) 1.51 m E) 1.89 m

Accepted Answer

To find molality, we need moles of solute per kg of solvent. In a 100 g solution, 20 g is ammonium sulfate, and 80 g is water. The molar mass of (NH4)2SO4 is 132.14 g/mol, so moles of solute = 20 g / 132.14 g/mol = 0.151 moles. The mass of water is 80 g = 0.080 kg. Molality = 0.151 moles / 0.080 kg = 1.89 m.

Question 15

What volume of a 2.75 M solution of NaOH is required to make 500.0 mL of a 1.27 M solution of NaOH?
A) 231 mL
B) 1.72 L
C) 1.10 L
D) 440 mL
E) 909 mL

Accepted Answer

To find the volume of the 2.75 M solution needed to make 500.0 mL of a 1.27 M solution, use the dilution formula $M_1V_1 = M_2V_2$, where $M_1$ and $V_1$ are the molarity and volume of the initial solution, and $M_2$ and $V_2$ are the molarity and volume of the final solution. Plugging in the given values: $2.75M \times V_1 = 1.27M \times 500.0mL$, solving for $V_1$ gives approximately 231 mL.

Question 16

What is the molality of a solution prepared by dissolving 84.7 g of KMnO₄ in 165 g of water? A) 0.339 m B) 0.513 m C) 0.536 m D) 3.25 m E) 81.1 m

Accepted Answer

Molality (m) is calculated by dividing the moles of solute by the kilograms of solvent. The molar mass of KMnO4 is approximately 158.04 g/mol. Thus, 84.7 g of KMnO4 is 0.536 moles. The mass of water is 165 g, which is 0.165 kg. Therefore, the molality is 0.536 moles / 0.165 kg = 3.25 m.

Question 17

Which of the following states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution?
A) Entropy
B) Henry's law
C) Dissolution
D) Vapor pressure
E) Enthalpy of solvation

Accepted Answer

Henry's law states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution. This means that as the pressure of the gas increases, the solubility of the gas in the liquid also increases.

Question 18

What mass of water is required to dissolve 27.8 g of ammonium nitrate NH₄NO₃ in order to prepare a 0.452 m solution? A) 0.0615 kg B) 0.100 kg C) 0.177 kg D) 0.768 kg E) 1.30 kg

Accepted Answer

The answer of What mass of water is required to...

Question 19

What is the molarity of a solution prepared by diluting 1.85 L of 6.5 M KOH to 11.0 L?
A) 0.28 M
B) 0.91 M
C) 1.1 M
D) 3.1 M
E) 3.9 M

Accepted Answer

The answer of What is the molarity of a solution...

Question 20

What mass of water is required to dissolve 25.31 g of potassium nitrate (KNO₃) in order to prepare a 0.1982 m solution? A) 0.5101 kg B) 0.1277 kg C) 1.000. kg D) 1.263 kg E) 0.7917 kg

Accepted Answer

The answer of What mass of water is required to...

Quiz 13: Physical Properties of Solutions

What mass of LiOH is required to prepare 0.250 L of a 3.55 M solution?

What volume of concentrated (14.7 M) phosphoric acid is needed to prepare 25.0 L of 3.0 M H₃PO₄?

Calcium nitrite is used as a corrosion inhibitor in lubricants. What is the molality of a solution prepared by dissolving 18.5 g of calcium nitrite in 83.5 g of distilled water?

The solubility of nitrogen gas in water at 25°C and a nitrogen pressure of 522 mmHg is 4.7 × 10^-4 mol/L. What is the value of the Henry's law constant in mol L^-1• atm^-1?

Cadmium bromide is used in photography and lithography. What is the molality of a solution prepared by dissolving 45.38 g of CdBr₂ in 375.0 g of water?

Copper(II) bromide is used as a wood preservative. What mass of CuBr₂ is needed to prepare 750.0 mL of a 1.25 M solution?

What is the molarity of a solution of 10.0% by mass cadmium sulfate, CdSO₄ (208.47 g/mol) ? The density of the solution is 1.10 g/mL.

The density of a 20.3 M CH₃OH (methanol) solution is 0.858 g/mL. What is the molality of this solution? H₂O is the solvent.

Which law describes the quantitative relationship between the solubility of a gas and its partial pressure?

At 10°C one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. What is the Henry's law constant of Cl₂ in water? (R = 0.0821 atm • L • mol^-1•K^-1 )

Which combination of a 36.0% (w/w) stock solution of acetic acid (d = 1.045 g/mL) and water will result in 1.00 kg of a 15.0% (w/w) acetic acid solution?

What is the molarity of a 17.0% by mass solution of sodium acetate, NaC₂H₃O₂ (82.0 g/mol) , in water? The density of the solution is 1.09 g/mL.

Calculate the molality of a 20.0% by mass ammonium sulfate (NH₄) ₂SO₄ solution. The density of the solution is 1.117 g/mL.

What volume of a 2.75 M solution of NaOH is required to make 500.0 mL of a 1.27 M solution of NaOH?

What is the molality of a solution prepared by dissolving 84.7 g of KMnO₄ in 165 g of water?

Which of the following states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution?

What mass of water is required to dissolve 27.8 g of ammonium nitrate NH₄NO₃ in order to prepare a 0.452 m solution?

What is the molarity of a solution prepared by diluting 1.85 L of 6.5 M KOH to 11.0 L?

What mass of water is required to dissolve 25.31 g of potassium nitrate (KNO₃) in order to prepare a 0.1982 m solution?

Quiz 13: Physical Properties of Solutions

What mass of LiOH is required to prepare 0.250 L of a 3.55 M solution?

What volume of concentrated (14.7 M) phosphoric acid is needed to prepare 25.0 L of 3.0 M H3PO4?

Calcium nitrite is used as a corrosion inhibitor in lubricants. What is the molality of a solution prepared by dissolving 18.5 g of calcium nitrite in 83.5 g of distilled water?

The solubility of nitrogen gas in water at 25°C and a nitrogen pressure of 522 mmHg is 4.7 × 10-4 mol/L. What is the value of the Henry's law constant in mol L-1• atm-1?

Cadmium bromide is used in photography and lithography. What is the molality of a solution prepared by dissolving 45.38 g of CdBr2 in 375.0 g of water?

Copper(II) bromide is used as a wood preservative. What mass of CuBr2 is needed to prepare 750.0 mL of a 1.25 M solution?

What is the molarity of a solution of 10.0% by mass cadmium sulfate, CdSO4 (208.47 g/mol) ? The density of the solution is 1.10 g/mL.

The density of a 20.3 M CH3OH (methanol) solution is 0.858 g/mL. What is the molality of this solution? H2O is the solvent.