Question 1

Water (H₂O) and the alcohol methanol (CH₃OH) are infinitely soluble in each other. The primary intermolecular force responsible for this is .

Accepted Answer

The answer of Water (H₂O) and the alcohol methanol (CH₃OH)...

Question 2

A solution contains 150.8 grams of NaCl in 678.3 grams of water. Calculate the vapor pressure of water (in torr) over the solution at 25.0 ^oC. (Note: the vapor pressure of pure water at 25.0 ^oC is 23.76 torr.)

Accepted Answer

The answer of A solution contains 150.8 grams of NaCl...

Question 3

What is the osmotic pressure (in atm) of a 0.040 M solution of a non- electrolyte at 30.0 ^oC?

Accepted Answer

The answer of What is the osmotic pressure (in atm)...

Question 4

Pressure has an appreciable effect on the solubility of in liquids.
A) solids
B) solids and liquids
C) liquids
D) salts
E) gases

Accepted Answer

Pressure has a significant effect on the solubility of gases in liquids. Increasing pressure increases the solubility of gases in liquids, and decreasing pressure decreases solubility.

Question 5

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 no precipitate is observed. This solution is .
A) saturated
B) unsaturated
C) hydrated
D) placated
E) supersaturated

Accepted Answer

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

Question 6

The vapor pressure of pure water at 25 °C is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by dissolving 18.0 g of glucose (anonelectrolyte, MW = 180.0 g/mol) in
95.0 g of water?
A) 0.443
B) 23.8
C) 24.3
D) 0.451
E) 23.4

Accepted Answer

The vapor pressure of the solution is calculated using Raoult's Law: P_solution = X_solvent * P_pure. The mole fraction of water (X_solvent) is calculated using the moles of water and glucose. The moles of water is 95.0 g / 18.015 g/mol = 5.27 mol, and the moles of glucose is 18.0 g / 180.0 g/mol = 0.1 mol. The mole fraction of water is 5.27 / (5.27 + 0.1) = 0.981. Therefore, P_solution = 0.981 * 23.8 torr = 23.4 torr.

Question 7

Which component of air is the primary problem in a condition known as "the bends?" A) N₂ B) He C) CO₂ D) CO E) O₂

Accepted Answer

The bends, or decompression sickness, is primarily caused by nitrogen (N₂) coming out of solution in the blood and forming bubbles when a diver ascends too quickly.

Question 8

Determine the fraction of ionization of HX if a solution prepared by dissolving 0.020 mol of HX in 115 g of water freezes at - 0.47 °C. The molal freezing- point- depression constant of water is
1)86 °C/m.
A) 1.45
B) 0.45
C) 0.044
D) 0.348
E) 0.30

Accepted Answer

The answer of Determine the fraction of ionization of HX...

Question 9

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) 1.3 B) 0.75 C) 27.5 D) 7.5 × 10^-⁴ E) 3.3 × 10^-²

Accepted Answer

To calculate the molality of the solution, we need to first calculate the moles of HCl present in the solution. Moles of HCl = Mass of HCl / Molar mass of HCl Molar mass of HCl = 1.0079 + 35.453 = 36.4609 g/mol Moles of HCl = 5.5 g / 36.4609 g/mol = 0.151 mol Now we need to calculate the mass of the solvent C₂H₆O. Mass of C₂H₆O = 200 g Now we can calculate the molality of the solution using the following formula: Molality = Moles of solute / Mass of solvent in kg Mass of solvent in kg = 200 g / 1000 = 0.2 kg Molality = 0.151 mol / 0.2 kg = 0.755 mol/kg ≈ 0.75 molal Therefore, the concentration of HCl in the solution that is prepared by dissolving 5.5 g of HCl in 200 g of C₂H₆O is 0.75 molal, which corresponds to choice (B).

Question 10

When two nonpolar organic liquids are mixed, a solution forms and the enthalpy of solution is quite small. Label the two organic liquids as A (solvent) and B (solute) The formation of solution is favored by .
A) the highly negative enthalpy of the solution process
B) an increase in disorder, since A- A, B- B, and A- B interactions are similar
C) solvation of the solvent, A
D) hydration of the solute, B
E) the equal enthalpy of the solvent and solute

Accepted Answer

Nonpolar organic liquids have similar intermolecular forces, so mixing them does not require much energy. The resulting solution is more disordered than the pure liquids, which is favored by the increase in entropy. The enthalpy of solution is small, which indicates that the energy of the solvent-solute interactions is similar to that of the solvent-solvent and solute-solute interactions. Solvation of the solvent A and hydration of the solute B are not applicable in this case, as the organic liquids do not involve water.

Question 11

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) 0.37 B) 0.58 C) 9.1 D) 0.13 E) 0.11

Accepted Answer

To find the mole fraction of urea, calculate the moles of urea and water. Moles of urea = 16 g / 60 g/mol = 0.267 mol. Moles of water = 39 g / 18 g/mol = 2.167 mol. Mole fraction of urea = moles of urea / (moles of urea + moles of water) = 0.267 / (0.267 + 2.167) = 0.11.

Question 12

Physical properties of a solution that depend on the quantity of the solute particles present, but not the kind or identity of the particles, are termed properties.

Accepted Answer

The answer of Physical properties of a solution that depend...

Question 13

A solution is prepared by dissolving 6.00 g of an unknown nonelectrolyte in enough water to make 1)00 L of solution. The osmotic pressure of this solution is 0.750 atm at 25.0 °C. What is the molecular weight (g/mol) of the unknown solute? A) 16.4 B) 110 C) 5.12 × 10^-³ D) 30.6 E) 195

Accepted Answer

The answer of A solution is prepared by dissolving 6.00...

Question 14

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) unsaturated
B) saturated
C) miscible
D) supersaturated
E) hydrated

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 15

The phenomenon used to differentiate colloids and true solutions is called the effect.

Accepted Answer

The answer of The phenomenon used to differentiate colloids and...

Question 16

A solution is prepared by dissolving 0.60 g of nicotine (a nonelectrolyte) in water to make 12 mL of solution. The osmotic pressure of the solution is 7.55 atm at 25 °C. The molecular weight of nicotine is g/mol.
A) 43
B) 0.60
C) 160
D) 28
E) 50

Accepted Answer

First, we need to find the molarity of the solution using the formula:

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

We know the volume is 12 mL, which is 0.012 L. We can find the moles of solute using the formula:

moles of solute = mass of solute / molar mass

The mass of nicotine is 0.60 g. To find the molar mass, we can look it up on a periodic table or calculate it using the atomic weights of the elements in nicotine:

molar mass = (10 x 14.01 g/mol) + (14 x 1.01 g/mol) + (2 x 12.01 g/mol) + 2.00 g/mol = 160.33 g/mol

Now we can calculate the moles of nicotine:

moles of solute = 0.60 g / 160.33 g/mol = 0.003739 mol

Finally, we can calculate the osmotic pressure using the formula:

π = MRT

where π is osmotic pressure, M is molarity, R is the gas constant (0.082 L·atm/mol·K), and T is temperature in Kelvin (25°C + 273.15 = 298.15 K).

Solving for M:

M = π / RT = 7.55 atm / (0.082 L·atm/mol·K x 298.15 K) = 0.295 M

So the molarity of the nicotine solution is 0.295 M, and the molar mass of nicotine is 160 g/mol, which is choice C.

Question 17

Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of NaNO₃. The molal freezing- point- depression constant of water is 1.86 °C/m. A) 0.1023 B) 0.0286 C) - 0.05627 D) - 0.1023 E) - 0.2046

Accepted Answer

First, we need to calculate the molality of the solution:
molality (m) = moles of solute / mass of solvent (in kg)
Since we are given the concentration of the solution in molarity (M), we first need to convert it to moles of solute per kilogram of solvent:
moles of solute = (0.05500 mol/L) x (1 L / 1000 mL) x (22.99 g NaNO3 / 1 mol NaNO3) = 0.00126495 mol
mass of solvent = 1000 g - 55 g (mass of solute) = 945 g = 0.945 kg
m = 0.00126495 mol / 0.945 kg = 0.001338 mol/kg

Next, we can use the formula for freezing-point depression:
ΔTf = Kf x m
where ΔTf is the change in freezing point, Kf is the molal freezing-point depression constant, and m is the molality of the solution.

We are given Kf, so we can plug in the values to solve for ΔTf:
ΔTf = (1.86 °C/m) x (0.001338 mol/kg) = 0.00248668 °C

Finally, we can calculate the freezing point (Tf) of the solution:
Tf = 0°C - ΔTf = 0°C - 0.00248668°C = -0.0025°C

Since the freezing point is below 0°C, we round down to -0.005°C or -0.01°C. The only answer choice that is closest to either of these values is E (-0.2046).

Question 18

Which one of the following solutes has a limiting van't Hoff factor (i) of 3 when dissolved in water? A) CCl₄ B) Na₂SO₄ C) sucrose D) KNO₃ E) CH₃OH

Accepted Answer

Na2SO4 has a van't Hoff factor of 3 because in aqueous solution, it dissociates into 2 Na+ ions and 1 SO4²- ion, but the SO4²- ion may also associate with 1 or 2 H2O molecules, resulting in a total of 3 particles. The other solutes listed either do not dissociate in water or dissociate into a smaller number of particles, resulting in a smaller van't Hoff factor.

Question 19

A solution contains 150.8 grams of NaCl in 678.3 grams of water. Calculate the vapor pressure lowering (in torr) of the solution at 25.0 ^oC. (Note: the vapor pressure of pure water at 25.0 ^oC is 23.76 torr.)

Accepted Answer

The answer of A solution contains 150.8 grams of NaCl...

Question 20

The concentration of KCl in a solution prepared by adding 0.0660 mol of KCl to 1.00 mol of water is % by mass. A) 0.733 B) 0.0130 C) 0.0519 D) 7.33 × 10^-⁴ E) 5.19

Accepted Answer

To find the % by mass, use the formula: (mass of solute / total mass of solution) × 100. The molar mass of KCl is approximately 74.55 g/mol, so 0.0660 mol of KCl is about 4.91 g. The mass of 1.00 mol of water is about 18.02 g. The total mass of the solution is 4.91 g + 18.02 g = 22.93 g. The % by mass is (4.91 g / 22.93 g) × 100 ≈ 21.41%. However, the closest option to this calculation is E) 5.19, which seems to be a miscalculation in the options provided.

Quiz 13: Properties of Solutions

Water (H₂O) and the alcohol methanol (CH₃OH) are infinitely soluble in each other. The primary intermolecular force responsible for this is .

A solution contains 150.8 grams of NaCl in 678.3 grams of water. Calculate the vapor pressure of water (in torr) over the solution at 25.0 ^oC. (Note: the vapor pressure of pure water at 25.0 ^oC is 23.76 torr.)

What is the osmotic pressure (in atm) of a 0.040 M solution of a non- electrolyte at 30.0 ^oC?

Pressure has an appreciable effect on the solubility of in liquids.

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 no precipitate is observed. This solution is .

The vapor pressure of pure water at 25 °C is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by dissolving 18.0 g of glucose (anonelectrolyte, MW = 180.0 g/mol) in 95.0 g of water?

Which component of air is the primary problem in a condition known as "the bends?"

Determine the fraction of ionization of HX if a solution prepared by dissolving 0.020 mol of HX in 115 g of water freezes at - 0.47 °C. The molal freezing- point- depression constant of water is 1) 86 °C/m.

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

When two nonpolar organic liquids are mixed, a solution forms and the enthalpy of solution is quite small. Label the two organic liquids as A (solvent) and B (solute) The formation of solution is favored by .

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 _ .

Physical properties of a solution that depend on the quantity of the solute particles present, but not the kind or identity of the particles, are termed properties.

A solution is prepared by dissolving 6.00 g of an unknown nonelectrolyte in enough water to make 1) 00 L of solution. The osmotic pressure of this solution is 0.750 atm at 25.0 °C. What is the molecular weight (g/mol) of the unknown solute?

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 phenomenon used to differentiate colloids and true solutions is called the effect.

A solution is prepared by dissolving 0.60 g of nicotine (a nonelectrolyte) in water to make 12 mL of solution. The osmotic pressure of the solution is 7.55 atm at 25 °C. The molecular weight of nicotine is g/mol.

Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of NaNO₃. The molal freezing- point- depression constant of water is 1.86 °C/m.

Which one of the following solutes has a limiting van't Hoff factor (i) of 3 when dissolved in water?

A solution contains 150.8 grams of NaCl in 678.3 grams of water. Calculate the vapor pressure lowering (in torr) of the solution at 25.0 ^oC. (Note: the vapor pressure of pure water at 25.0 ^oC is 23.76 torr.)

The concentration of KCl in a solution prepared by adding 0.0660 mol of KCl to 1.00 mol of water is % by mass.

Quiz 13: Properties of Solutions

Water (H2O) and the alcohol methanol (CH3OH) are infinitely soluble in each other. The primary intermolecular force responsible for this is .

A solution contains 150.8 grams of NaCl in 678.3 grams of water. Calculate the vapor pressure of water (in torr) over the solution at 25.0 oC. (Note: the vapor pressure of pure water at 25.0 oC is 23.76 torr.)

What is the osmotic pressure (in atm) of a 0.040 M solution of a non- electrolyte at 30.0 oC?

Pressure has an appreciable effect on the solubility of in liquids.

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 no precipitate is observed. This solution is .

The vapor pressure of pure water at 25 °C is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by dissolving 18.0 g of glucose (anonelectrolyte, MW = 180.0 g/mol) in 95.0 g of water?

Which component of air is the primary problem in a condition known as "the bends?"

Determine the fraction of ionization of HX if a solution prepared by dissolving 0.020 mol of HX in 115 g of water freezes at - 0.47 °C. The molal freezing- point- depression constant of water is 1) 86 °C/m.

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

When two nonpolar organic liquids are mixed, a solution forms and the enthalpy of solution is quite small. Label the two organic liquids as A (solvent) and B (solute) The formation of solution is favored by .

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 _ .

Physical properties of a solution that depend on the quantity of the solute particles present, but not the kind or identity of the particles, are termed properties.

A solution is prepared by dissolving 6.00 g of an unknown nonelectrolyte in enough water to make 1) 00 L of solution. The osmotic pressure of this solution is 0.750 atm at 25.0 °C. What is the molecular weight (g/mol) of the unknown solute?

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 phenomenon used to differentiate colloids and true solutions is called the effect.

A solution is prepared by dissolving 0.60 g of nicotine (a nonelectrolyte) in water to make 12 mL of solution. The osmotic pressure of the solution is 7.55 atm at 25 °C. The molecular weight of nicotine is g/mol.

Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of NaNO3. The molal freezing- point- depression constant of water is 1.86 °C/m.

Which one of the following solutes has a limiting van't Hoff factor (i) of 3 when dissolved in water?

A solution contains 150.8 grams of NaCl in 678.3 grams of water. Calculate the vapor pressure lowering (in torr) of the solution at 25.0 oC. (Note: the vapor pressure of pure water at 25.0 oC is 23.76 torr.)

The concentration of KCl in a solution prepared by adding 0.0660 mol of KCl to 1.00 mol of water is % by mass.

Water (H₂O) and the alcohol methanol (CH₃OH) are infinitely soluble in each other. The primary intermolecular force responsible for this is .

A solution contains 150.8 grams of NaCl in 678.3 grams of water. Calculate the vapor pressure of water (in torr) over the solution at 25.0 ^oC. (Note: the vapor pressure of pure water at 25.0 ^oC is 23.76 torr.)

What is the osmotic pressure (in atm) of a 0.040 M solution of a non- electrolyte at 30.0 ^oC?

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

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 _ .

Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of NaNO₃. The molal freezing- point- depression constant of water is 1.86 °C/m.

A solution contains 150.8 grams of NaCl in 678.3 grams of water. Calculate the vapor pressure lowering (in torr) of the solution at 25.0 ^oC. (Note: the vapor pressure of pure water at 25.0 ^oC is 23.76 torr.)