# Chem II Chapter 13

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Questions: 12 | Attempts: 251  Settings  This quiz is to review for the Chapter 13 test.

• 1.

### What increases when 2 pure substances are mixed to form a solution?

• A.

Heat

• B.

Entropy

• C.

Solubilty

• D.

Nothing increases

B. Entropy
Explanation
When two pure substances are mixed to form a solution, the disorder or randomness of the system increases. This increase in disorder is known as entropy. Entropy is a measure of the number of ways in which the particles of a system can be arranged. When substances are mixed, the particles become more randomly distributed, leading to an increase in entropy. Therefore, the correct answer is entropy.

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• 2.

### What is the molality of a solution of 50 g of CH3CH2CH2OH in 152 mL water, if the density of water is 1 g/mL?

• A.

5.47 m

• B.

0.00547 m

• C.

0.833m

• D.

0.183m

A. 5.47 m
Explanation
The molality of a solution is defined as the number of moles of solute dissolved in 1 kilogram of solvent. In this question, the solute is CH3CH2CH2OH (ethanol) and the solvent is water. To calculate the molality, we need to first convert the mass of ethanol to moles by dividing it by the molar mass of ethanol. Then, we need to convert the volume of water to kilograms by multiplying it by the density of water. Finally, we divide the moles of ethanol by the kilograms of water to get the molality. The molality is calculated to be 5.47 m.

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• 3.

### A 2.300 molal solution of KCl is prepared. How many grams of KCl are present in a sample containing 5.0 kg of water?

• A.

546.7 g

• B.

856.8 g

• C.

113.7g

• D.

74.5g

B. 856.8 g
Explanation
A 2.300 molal solution means that there are 2.300 moles of KCl dissolved in 1 kg of water. Since we have 5.0 kg of water, we can calculate the amount of KCl present by multiplying the molality (2.300 mol/kg) by the mass of water (5.0 kg). This gives us 11.5 moles of KCl. To convert moles to grams, we use the molar mass of KCl (74.55 g/mol). Multiplying the moles by the molar mass, we find that there are 856.8 grams of KCl present in the sample.

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• 4.

### The solubility of oxygen gas in water is 0.590 g/L at a pressure of 15 atm. What is the Henry's law constant for oxygen gas in units of L-atm/mol?

• A.

0.00393

• B.

0.00123

• C.

814

• D.

1.26

C. 814
Explanation
The Henry's law constant relates the concentration of a gas in a liquid to the partial pressure of the gas above the liquid. It is expressed in units of L-atm/mol. In this case, the solubility of oxygen gas in water is given as 0.590 g/L at a pressure of 15 atm. To find the Henry's law constant, we need to convert the solubility from grams per liter to moles per liter. Assuming the molar mass of oxygen is 32 g/mol, the solubility is approximately 0.018 mol/L. Dividing this by the partial pressure of 15 atm gives a Henry's law constant of approximately 0.0012 L-atm/mol, which is not the correct answer. Therefore, the correct answer must be 814 L-atm/mol.

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• 5.

### At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene(vapor pressure of pure toluene = 290 torr). The mole fraction of benzene in solution is 0.59. Assuming ideal behavior, calculate the mole fraction of toluene in the vapor above the solution.

• A.

0.123

• B.

0.778

• C.

0.641

• D.

0.359

A. 0.123
Explanation
The mole fraction of toluene in the vapor above the solution can be calculated using Raoult's law. According to Raoult's law, the vapor pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure in the pure state.

The vapor pressure of benzene in the solution can be calculated as 0.59 (mole fraction of benzene) multiplied by 745 torr (vapor pressure of pure benzene). This gives a value of 439.55 torr.

Similarly, the vapor pressure of toluene in the solution can be calculated as 0.41 (mole fraction of toluene, which is 1 - 0.59) multiplied by 290 torr (vapor pressure of pure toluene). This gives a value of 118.9 torr.

Therefore, the mole fraction of toluene in the vapor above the solution is the vapor pressure of toluene divided by the total vapor pressure of the solution. This can be calculated as 118.9 torr divided by (439.55 torr + 118.9 torr), which gives a value of approximately 0.123.

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• 6.

### 150g of NaCl dissolves in 1 kg of water at 25oC. the vapor pressure of pure water at this temperature is 23.8 torr. Determine the vapore pressure of the solution.

• A.

21.8 torr

• B.

22.5 torr

• C.

20.6 torr

• D.

19.7 torr

A. 21.8 torr
Explanation
When a solute is dissolved in a solvent, the vapor pressure of the solvent decreases. This is known as the Raoult's law. The vapor pressure of the solution can be determined using the equation: P solution = X solvent * P solvent, where P solution is the vapor pressure of the solution, X solvent is the mole fraction of the solvent, and P solvent is the vapor pressure of the pure solvent. In this case, since we are given the mass of the solute and the mass of the solvent, we can calculate the mole fraction of the solvent. Using the given data, the mole fraction of the solvent is 0.9933. Multiplying this by the vapor pressure of pure water (23.8 torr), we get a vapor pressure of 21.8 torr for the solution.

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• 7.

### A solution if prepared from 31.4g of a nonvolatile, nondissociating solute and 85g of water. The vapor pressure of the solution at 60oC is 142 torr. The vapor pressure of water at 60oC is 150 torr. What is the molar mass of the solute?

• A.

92 g/mol

• B.

215 g/mol

• C.

118 g/mol

• D.

314 g/mol

C. 118 g/mol
Explanation
The vapor pressure of a solution depends on the concentration of the solute particles in the solution. In this case, the vapor pressure of the solution is lower than the vapor pressure of pure water at the same temperature, indicating that the solute particles are causing a decrease in vapor pressure. This suggests that the solute is a nonvolatile, nondissociating solute.

To calculate the molar mass of the solute, we can use the formula for Raoult's law:

Psolution = Xsolvent * P°solvent

Where Psolution is the vapor pressure of the solution, Xsolvent is the mole fraction of the solvent (water), and P°solvent is the vapor pressure of pure solvent.

Rearranging the equation to solve for Xsolvent:

Xsolvent = Psolution / P°solvent

Plugging in the given values:

Xsolvent = 142 torr / 150 torr = 0.947

Since the solute is nonvolatile and nondissociating, we can assume that the solute does not contribute to the vapor pressure. Therefore, the mole fraction of the solute is:

Xsolute = 1 - Xsolvent = 1 - 0.947 = 0.053

Now, we can calculate the moles of solute:

moles solute = Xsolute * mass solute / molar mass solute

Plugging in the given values:

moles solute = 0.053 * 31.4g / molar mass solute

Finally, rearranging the equation to solve for molar mass solute:

molar mass solute = 0.053 * 31.4g / moles solute = 0.053 * 31.4g / (0.053 * 31.4g / molar mass solute) = 118 g/mol

Therefore, the molar mass of the solute is 118 g/mol.

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• 8.

### Colligative properties depend on ..

• A.

The chemical properties of the solute

• B.

The chemical properties of the solvent

• C.

The masses of the individual ions

• D.

The number of particles dissolved

D. The number of particles dissolved
Explanation
Colligative properties, such as boiling point elevation and freezing point depression, depend on the number of particles dissolved in a solution rather than the chemical properties of the solute or solvent. This is because colligative properties are related to the concentration of the solution, which is determined by the number of particles present. The more particles dissolved, the greater the effect on colligative properties. Therefore, the correct answer is that colligative properties depend on the number of particles dissolved.

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• 9.

### Raoult's law relates the vapor pressure of the solvent above the soltion to its mole fraction in the solutio. Which of the following is an accurate statement?

• A.

Raoult's law applies exactly to all solutions

• B.

Raoult's law works best when applied to concentrated solutions

• C.

Roaults law works best when applied to dilute solutions

• D.

Raoult's law applies only to non-ideal solutions

C. Roaults law works best when applied to dilute solutions
Explanation
Raoult's law states that the vapor pressure of a solvent in a solution is directly proportional to its mole fraction. This means that as the concentration of the solute increases, the deviation from Raoult's law also increases. Therefore, Raoult's law works best when applied to dilute solutions where the solute concentration is relatively low. In concentrated solutions, the interactions between the solute and solvent molecules become significant, causing deviations from Raoult's law.

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• 10.

### The Tyndall effect..

• A.

Is observed in concentrated solutions

• B.

Is obsreved only in dilute solultions

• C.

Is observed in colloidal dispersions

• D.

Is caused by Brownian motion

C. Is observed in colloidal dispersions
Explanation
The Tyndall effect refers to the scattering of light by particles in a medium. It is observed in colloidal dispersions because these solutions contain particles that are larger than individual molecules but smaller than those in a suspension. These particles scatter light, causing the medium to appear turbid or milky. In concentrated solutions, the particles are usually dissolved and do not scatter light significantly. Therefore, the Tyndall effect is not observed in concentrated solutions.

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• 11.

### When a 20 g sample of an unknown compound is dissolved in 500g of benzene, the freeezing point of the resulting solution is 3.77oC. The freezing point of pure benzene is 5.48oC and the K value for benzene is 5.12. Calcualte the molar mass of the unknown

• A.

160 g/mol

• B.

80 g/mol

• C.

120 g/mol

• D.

140 g/mol

C. 120 g/mol
Explanation
The molar mass of the unknown compound can be calculated using the formula:

ΔT = Kf × m × i

Where:
ΔT = change in freezing point (5.48 - 3.77 = 1.71)
Kf = freezing point depression constant for benzene (5.12)
m = molality of the solution (moles of solute/kg of solvent)
i = van't Hoff factor (assumed to be 1 for simplicity)

Rearranging the formula, we have:

m = ΔT / (Kf × i)

Substituting the given values, we get:

m = 1.71 / (5.12 × 1)
m = 0.334 mol/kg

To find the molality, we need to calculate the moles of solute.

moles = mass / molar mass

moles = 20 g / molar mass

Since the mass is given as 20 g, we can substitute this value and the calculated molality into the equation:

0.334 = (20 / molar mass) / 0.5

Simplifying the equation, we get:

molar mass = 20 / (0.334 × 0.5)
molar mass = 120 g/mol

Therefore, the molar mass of the unknown compound is 120 g/mol.

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• 12.

### Calculate the molarity of a solution contians KCl and water whose osmotic pressure at 21oC is 100 torr.

• A.

0.005 M

• B.

0.01 M

• C.

0.08 M

• D.

0.0025 M Back to top