Question 1

The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr) of water at 25 °C above a solution prepared by dissolving 35 g of urea (a nonvolatile, non- electrolyte, MW =60.0 g/mol) in 75 g of water.
A) 0.88
B) 27
C) 21
D) 3.3
E) 2.9

Accepted Answer

First, we need to calculate the mole fraction of water and urea in the solution.

Moles of solute (urea) = mass/molar mass = 35/60.0 = 0.5833 mol
Moles of solvent (water) = mass/molar mass = 75/18.0 = 4.17 mol

Total moles in solution = moles of solute + moles of solvent = 4.17 + 0.5833 = 4.7533 mol

Mole fraction of water = moles of water / total moles = 4.17 / 4.7533 = 0.8773
Mole fraction of urea = moles of urea / total moles = 0.5833 / 4.7533 = 0.1227

Now we can use Raoult's law to calculate the vapor pressure of the solution.

P solution = X water * P water

where P water is the vapor pressure of pure water and X water is the mole fraction of water in the solution.

P solution = 0.8773 * 23.8 torr = 20.9 torr

Finally, we use the equation for the vapor pressure lowering (ΔP = P° - P solution) to find the vapor pressure of water above the solution.

ΔP = P° - P solution
ΔP = 23.8 torr - 20.9 torr
ΔP = 2.9 torr

Therefore, the vapor pressure (torr) of water at 25 °C above the solution is 2.9 torr. The best answer choice is C.

Question 2

On a clear day at sea level, with a temperature of 25 °C, the partial pressure of N₂in air is 0.78 atm and the concentration of nitrogen in water is 5.3 × 10^-⁴M. When the partial pressure of N₂is Atm, the concentration in water is 1.1 × 10^-³M. A) 1.6 atm B) 0.78 atm C) 2.1 atm D) 0.63 atm E) 1.0 atm

Accepted Answer

According to Henry's law, the concentration of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Therefore, we can set up a proportion:

concentration at 1.6 atm / concentration at 0.78 atm = partial pressure at 1.6 atm / partial pressure at 0.78 atm

Solving for the concentration at 1.6 atm, we get:

concentration at 1.6 atm = concentration at 0.78 atm * (partial pressure at 1.6 atm / partial pressure at 0.78 atm)

Plugging in the given values, we get:

concentration at 1.6 atm = (1.1 × 10^-13 M) * (1.6 atm / 0.78 atm) = 2.3 × 10^-13 M

Therefore, the concentration of nitrogen in water when the partial pressure is 1.6 atm is 2.3 × 10^-13 M, which corresponds to choice A).

Question 3

Which one of the following vitamins is water soluble?
A) A
B) B
C) K
D) D
E) E

Accepted Answer

Vitamins B and C are the only water-soluble vitamins. Vitamins A, D, E, and K are fat-soluble vitamins.

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) 2.23 B) 143 C) 2.45 D) 2.56 E) 1.43 × 10^-²

Accepted Answer

First, we need to calculate the mass of the solution. Let's assume we have 100 g of the solution:
- 13.0 g of that is sodium chloride
- The rest (100 g - 13.0 g = 87.0 g) is water

Now, we can use the density of the solution to find the volume of the solution:
- density = mass/volume
- volume = mass/density
- volume = 100 g / 1.10 g/mL = 90.9 mL

Next, we can use the molarity equation to find the molarity of the sodium chloride:
- molarity = moles of solute / volume of solution (in L)
- To find the moles of solute, we need to convert the mass of sodium chloride to moles:
   - moles = mass / molar mass
   - molar mass of NaCl = 58.44 g/mol
   - moles of NaCl = 13.0 g / 58.44 g/mol = 0.222 mol
- Now we can plug in the values:
   - molarity = 0.222 mol / 0.0909 L = 2.45 M

Therefore, the answer is choice C.

Question 5

A solution is prepared by dissolving 2.60 g of a strong electrolyte (formula weight = 101 g/mol) in enough water to make 1.00 L of solution. The osmotic pressure of the solution is 1.25 atm at 25.0 °C. What is the van't Hoff factor (i) for the unknown solute?
A) 0
B) 0.99
C) 1.98
D) 2.98
E) 0.630

Accepted Answer

The answer of A solution is prepared by dissolving 2.60...

Question 6

The concentration of KBr in a solution prepared by dissolving 2.21 g of KBr in 897 g of water is Molal. A) 0.0207 B) 0.0167 C) 2.46 D) 0.0186 E) 2.07 × 10^-⁵

Accepted Answer

Molality = (Moles of solute / Kilograms of solvent)Moles of KBr = 2.21 g / 119 g/mol = 0.0186 molKilograms of water = 897 g / 1000 g/kg = 0.897 kgMolality = 0.0186 mol / 0.897 kg = 0.0207 mol/kg

Question 7

The concentration of a benzene solution prepared by mixing 12.0 g C₆H₆with 38.0 g CCl₄is Molal. A) 0.240 B) 4.04 C) 0.508 D) 0.622 E) 0.316

Accepted Answer

Molality is calculated by dividing the moles of solute by the kilograms of solvent. The moles of benzene (C6H6) is 12.0 g / 78.11 g/mol = 0.1537 mol. The mass of CCl4 in kg is 38.0 g / 1000 = 0.038 kg. Molality = 0.1537 mol / 0.038 kg = 4.04 mol/kg.

Question 8

A solution is prepared by dissolving 23.7 g of CaCl₂in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl₂is % by mass. A) 6.24 B) 0.0632 C) 0.0594 D) 5.94 E) 6.32

Accepted Answer

The mass percent of CaCl2 is calculated by dividing the mass of CaCl2 by the total mass of the solution and multiplying by 100. The total mass of the solution is 23.7 g + 375 g = 398.7 g. Therefore, the mass percent is (23.7 g / 398.7 g) * 100 = 5.94%.

Question 9

The concentration of urea in a solution prepared by dissolving 16 g of urea in 39 g of H₂O is % by mass. The molar mass of urea is 60.0 g/mol. A) 0.29 B) 0.41 C) 29 D) 0.48 E) 41

Accepted Answer

The answer of The concentration of urea in a solution...

Question 10

The solubility of oxygen gas in water at 25 °C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at 3.0 atm and 25 °C is g/L.
A) 3.0
B) 0.12
C) 0.31
D) 0.014
E) 0.041

Accepted Answer

According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. If the pressure is increased from 1.0 atm to 3.0 atm, the solubility will also increase by the same factor. Therefore, 0.041 g/L * 3 = 0.123 g/L, which is approximately 0.12 g/L.

Question 11

The solubility of MnSO₄monohydrate 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 MnSO₄monohydrate is best described as a(n) solution. The formula weight of MnSO₄monohydrate is 168.97 g/mol. A) solvated B) supersaturated C) hydrated D) unsaturated E) saturated

Accepted Answer

The answer of The solubility of MnSO₄monohydrate in water...

Question 12

A solution is prepared by dissolving 23.7 g of CaCl₂in 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) 1.20 B) 6.64 × 10^-² C) 1.12 D) 0.214 E) 0.562

Accepted Answer

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

Question 13

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) 0.68 B) 0.11 C) 6.9 D) 96 E) 6.3

Accepted Answer

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

Question 14

Pairs of liquids that will mix in all proportions are called liquids.
A) miscible
B) saturated
C) unsaturated
D) polar liquids
E) supersaturated

Accepted Answer

Miscible liquids are liquids that are able to mix in all proportions because they have similar intermolecular forces, such as polarity.

Question 15

A solution is prepared by dissolving 15.0 g of NH₃in 250 g of water. The density of the resulting solution is 0.974 g/mL. The molarity of NH₃in the solution is _ _. A) 3.53 B) 3.24 C) 0.882 D) 60.0 E) 0.00353

Accepted Answer

To find the molarity, first calculate the moles of NH3: 15.0 g / 17.03 g/mol = 0.880 moles. The total mass of the solution is 265 g (15 g NH3 + 250 g water). Using the density, the volume of the solution is 265 g / 0.974 g/mL = 272.2 mL = 0.2722 L. Molarity = moles/volume = 0.880 moles / 0.2722 L = 3.24 M.

Question 16

The solubility of Ar in water at 25 °C is 1.6 × 10^-³M when the pressure of the Ar above the solution is 1.0 atm. The solubility of Ar at a pressure of 2.5 atm is M. A) 4.0 × 10^-³ B) 7.5 × 10^-² C) 1.6 × 10^-³ D) 6.4 × 10^-⁴ E) 1.6 × 10³

Accepted Answer

According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. Therefore, if the pressure increases from 1.0 atm to 2.5 atm, the solubility will increase by the same factor. Thus, 1.6 × 10^-3 M × 2.5 = 4.0 × 10^-3 M.

Question 17

George is making spaghetti for dinner. He places 4.01 kg of water in a pan and brings it to a boil. Before adding the pasta, he adds 58 g of table salt to the water and again brings it to a boil. The temperature of the salty, boiling water is °C. It is a nice day at sea level so that pressure is 1.00 atm. Assume negligible evaporation of water. K_bfor water is 0.52°C/m. A) 100.13 B) 100.00 C) 100.26 D) 99.74 E) 99.87

Accepted Answer

First, we need to calculate the boiling point elevation due to the addition of salt using the formula ΔTb = Kb * molality of salt.

Since George added 58 g of table salt to 4.01 kg of water, the molality of salt is:

molality = moles of solute / mass of solvent (in kg)
moles of salt = 58 g / 58.44 g/mol = 0.992 mol
mass of water = 4.01 kg
molality = 0.992 mol / 4.01 kg = 0.247 mol/kg

Now, we can calculate the boiling point elevation:

ΔTb = Kb * molality of salt
Kb for water = 0.52°C/m
ΔTb = 0.52°C/m * 0.247 mol/kg = 0.128°C

The boiling point elevation is 0.128°C.

The normal boiling point of water is 100°C, so the new boiling point is:

new boiling point = normal boiling point + ΔTb
new boiling point = 100°C + 0.128°C = 100.128°C

Round to three decimal places: 100.26°C.

Therefore, the best choice is C, 100.26.

Question 18

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) 3.7 B) 6.8 C) 0.11 D) 6.3 E) 0.16

Accepted Answer

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

moles of urea = mass / molar mass = 16 g / 60.0 g/mol = 0.27 mol

Next, we need to calculate the volume of the solution. We can use the density to do this:

density = mass / volume

volume = mass / density = 39 g / 1.3 g/mL = 30 mL

Finally, we can calculate the molarity of the solution:

Molarity = moles of solute / volume of solution in liters

Molarity = 0.27 mol / (30 mL / 1000 mL/L) = 9.0 mM = 6.3 M

Therefore, the correct answer is choice D, 6.3.

Question 19

Determine the freezing point (°C) of a 0.015 molal aqueous solution of MgSO₄. Assume i = 2.0 for MgSO₄. The molal freezing- point- depression constant of water is 1.86 °C/m. A) 0.000 B) - 0.056 C) - 0.028 D) - 0.084 E) - 0.17

Accepted Answer

First, we need to use the van't Hoff factor (i) of 2.0 for MgSO4 to calculate the freezing point depression:

ΔTf = i * Kf * molality

where ΔTf = change in freezing point, Kf = molal freezing-point depression constant of water, and molality = moles of solute per kg of solvent.

Plugging in the values, we get:

ΔTf = 2.0 * 1.86 °C/m * 0.015 mol/kg = 0.056 °C

Since we're looking for the freezing point of the solution, we need to subtract this value from the freezing point of pure water, which is 0 °C. Therefore:

Freezing point = 0 °C - 0.056 °C = -0.056 °C

So the answer is B (-0.056).

Question 20

The concentration of lead nitrate (Pb(NO₃)₂) in a 0.726 M solution is molal. The density of the solution is 1.202 g/mL. A) 1.928 B) 0.650 C) 0.755 D) 0.819 E) 0.476

Accepted Answer

To find molality, use the formula: molality = moles of solute / kg of solvent. First, calculate the mass of the solution using its density (1.202 g/mL) and volume (1 L = 1000 mL). Then, find the mass of the solute (Pb(NO3)2) using its molarity (0.726 M) and molar mass. Subtract the mass of the solute from the total mass to get the mass of the solvent in kg. Finally, divide the moles of solute by the mass of the solvent in kg to find the molality.

Quiz 13: Properties of Solutions

The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr) of water at 25 °C above a solution prepared by dissolving 35 g of urea (a nonvolatile, non- electrolyte, MW =60.0 g/mol) in 75 g of water.

On a clear day at sea level, with a temperature of 25 °C, the partial pressure of N₂in air is 0.78 atm and the concentration of nitrogen in water is 5.3 × 10^-⁴M. When the partial pressure of N₂is Atm, the concentration in water is 1.1 × 10^-³M.

Which one of the following vitamins is water soluble?

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 solution is prepared by dissolving 2.60 g of a strong electrolyte (formula weight = 101 g/mol) in enough water to make 1.00 L of solution. The osmotic pressure of the solution is 1.25 atm at 25.0 °C. What is the van't Hoff factor (i) for the unknown solute?

The concentration of KBr in a solution prepared by dissolving 2.21 g of KBr in 897 g of water is Molal.

The concentration of a benzene solution prepared by mixing 12.0 g C₆H₆with 38.0 g CCl₄is Molal.

A solution is prepared by dissolving 23.7 g of CaCl₂in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl₂is % by mass.

The concentration of urea in a solution prepared by dissolving 16 g of urea in 39 g of H₂O is % by mass. The molar mass of urea is 60.0 g/mol.

The solubility of oxygen gas in water at 25 °C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at 3.0 atm and 25 °C is g/L.

The solubility of MnSO₄monohydrate 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 MnSO₄monohydrate is best described as a(n) solution. The formula weight of MnSO₄monohydrate is 168.97 g/mol.

A solution is prepared by dissolving 23.7 g of CaCl₂in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of Cl^-in this solution is _ M.

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.

Pairs of liquids that will mix in all proportions are called liquids.

A solution is prepared by dissolving 15.0 g of NH₃in 250 g of water. The density of the resulting solution is 0.974 g/mL. The molarity of NH₃in the solution is _ _.

The solubility of Ar in water at 25 °C is 1.6 × 10^-³M when the pressure of the Ar above the solution is 1.0 atm. The solubility of Ar at a pressure of 2.5 atm is M.

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.

Determine the freezing point (°C) of a 0.015 molal aqueous solution of MgSO₄. Assume i = 2.0 for MgSO₄. The molal freezing- point- depression constant of water is 1.86 °C/m.

The concentration of lead nitrate (Pb(NO₃) ₂) in a 0.726 M solution is molal. The density of the solution is 1.202 g/mL.

Quiz 13: Properties of Solutions

The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr) of water at 25 °C above a solution prepared by dissolving 35 g of urea (a nonvolatile, non- electrolyte, MW =60.0 g/mol) in 75 g of water.

On a clear day at sea level, with a temperature of 25 °C, the partial pressure of N2 in air is 0.78 atm and the concentration of nitrogen in water is 5.3 × 10- 4 M. When the partial pressure of N2 is Atm, the concentration in water is 1.1 × 10- 3 M.

Which one of the following vitamins is water soluble?

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 solution is prepared by dissolving 2.60 g of a strong electrolyte (formula weight = 101 g/mol) in enough water to make 1.00 L of solution. The osmotic pressure of the solution is 1.25 atm at 25.0 °C. What is the van't Hoff factor (i) for the unknown solute?

The concentration of KBr in a solution prepared by dissolving 2.21 g of KBr in 897 g of water is Molal.

The concentration of a benzene solution prepared by mixing 12.0 g C6H6 with 38.0 g CCl4 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 CaCl2 is % by mass.

The concentration of urea in a solution prepared by dissolving 16 g of urea in 39 g of H2O is % by mass. The molar mass of urea is 60.0 g/mol.

The solubility of oxygen gas in water at 25 °C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at 3.0 atm and 25 °C is g/L.

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 concentration of Cl- in this solution is _ M.

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.