Question 1

Arrange the following compounds in order of increasing boiling point: I.1-propanol,CH₃CH₂CH₂OH; II.1,2-propanediol,CH₃CH(OH)CH₂OH; III.1,2,3-propanetriol, (glycerol),HOCH₂CH(OH)CH₂OH.

Accepted Answer

The boiling point of compounds depends on their intermolecular forces. The strength of intermolecular forces increases with increasing polarity and increasing number of hydrogen bonds. 
Among the given compounds, 1-propanol can form hydrogen bonding with itself due to the presence of the hydroxyl (-OH) group. In 1,2-propanediol, two hydroxyl groups are present which can form more extensive hydrogen bonding than 1-propanol. In 1,2,3-propanetriol or glycerol, three hydroxyl groups are present, which can form even more extensive hydrogen bonding than 1,2-propanediol. 
Therefore, the increasing order of boiling points is I < II < III. 
Hence, the correct answer is choice B.

Question 2

A substance has a heat of fusion of 61.5 kJ/mol and a heat of deposition of -167.4 kJ/mol.What is the heat of sublimation in kJ/mol?

Accepted Answer

The answer of A substance has a heat of fusion...

Question 3

If a substance has a heat of condensation of -1.46 kJ/g and a heat of sublimation of 4.60 kJ/g,what is its heat of solidification in kJ/g?

Accepted Answer

The answer of If a substance has a heat of...

Question 4

If 34 g of a solid with a molar mass = 174 g/mol requires 21.3 kJ to melt it,what is the molar heat of fusion in kJ/mol?

Accepted Answer

The molar heat of fusion is the energy required to melt one mole of a substance, which can be calculated using the equation:

molar heat of fusion = energy required to melt / number of moles melted

First, we need to find the number of moles melted:

number of moles = mass / molar mass

number of moles = 34 g / 174 g/mol

number of moles = 0.195402

Next, we can use the given energy and number of moles to calculate the molar heat of fusion:

molar heat of fusion = 21.3 kJ / 0.195402 mol

molar heat of fusion = 108.947 kJ/mol

Therefore, the correct answer is D, (21.3)(174/34) kJ/mol.

Question 5

Given the data below,determine the normal boiling point of COCl₂. P₁ = 100 mmHg,t₁ = -35.6 °C,Δ_vapH = 27.4 kJ/mol

Accepted Answer

The normal boiling point of a substance is the temperature at which its vapor pressure equals 1 atm (760 mmHg). To find this, we can use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the enthalpy of vaporization. Given the initial vapor pressure (100 mmHg) and temperature (-35.6 °C), along with the enthalpy of vaporization (27.4 kJ/mol), we can solve for the temperature at which the vapor pressure equals 760 mmHg. The correct answer, after performing the calculations, is 5.0 °C, indicating that the normal boiling point of the substance is 5.0 °C.

Question 6

How much heat would be released by the condensation of 5.40 g of steam at 100 °C and the subsequent cooling of the water to 25 °C? [Δ_vapH = 40.7 kJ/mol at 100 °C;C_p for H₂O(l)is 4.18 J g^-1 °C^-1]

Accepted Answer

First, we need to calculate the amount of heat released during the condensation of 5.40 g of steam at 100 °C:

n(H2O) = 5.40 g / 18.02 g/mol = 0.299 mol

ΔH1 = n(H2O) x ΔHvap = 0.299 mol x 40.7 kJ/mol = 12.2 kJ

Next, we need to calculate the amount of heat released during the cooling of the water from 100 °C to 25 °C:

ΔT = 100 °C - 25 °C = 75 °C

q2 = m(H2O) x C(H2O) x ΔT = 5.40 g x 4.18 J/g°C x 75 °C = 1390.4 J = 1.3904 kJ

Therefore, the total amount of heat released is:

qtotal = ΔH1 + q2 = 12.2 kJ + 1.3904 kJ = 13.5904 kJ ≈ 13.9 kJ

So, the answer is choice C.

Question 7

A liquid has a molar heat of vaporization of 22.7 kJ/mol.Its normal boiling point is 459 K.What is the vapor pressure,in mmHg,at 70 °C?

Accepted Answer

The Clausius-Clapeyron equation can be used to calculate the vapor pressure at a new temperature given the heat of vaporization and the normal boiling point. The equation is ln(P1/P2) = ΔHvap/R * (1/T2 - 1/T1), where ΔHvap is the molar heat of vaporization, R is the gas constant (8.314 J/mol·K), T1 is the normal boiling point in Kelvin, and T2 is the new temperature in Kelvin. Converting 70°C to Kelvin gives 343K. Since the normal boiling point is 459K, where the vapor pressure equals 760 mmHg (standard atmospheric pressure), we can solve for P2, the vapor pressure at 343K. The calculation shows that the vapor pressure at 70°C is closest to 102 mmHg, making option A correct.

Question 8

When the vapor pressure of a liquid equals atmospheric pressure,the temperature of the liquid equals ________.

Accepted Answer

When the vapor pressure of a liquid equals atmospheric pressure, the liquid boils and its temperature reaches the boiling point. Therefore, choice B (the boiling point) is the best answer. Choice C (the normal boiling point) is not necessarily correct as the normal boiling point is the boiling point at standard conditions (1 atm pressure). Choice A (100°C) is not always true as the boiling point of a liquid varies with the atmospheric pressure. Choice D (the vaporization point) is not precise as it does not specify whether it refers to the boiling point or another point of vaporization. Choice E (the sublimation point) is not relevant as sublimation refers to the direct transition of a solid to a gas without passing through the liquid phase.

Question 9

The maximum temperature at which a gas can be liquefied is just below the ________.

Accepted Answer

The critical temperature is the highest temperature at which a gas can be liquefied. Above the critical temperature, it is not possible to liquefy a gas regardless of the amount of pressure applied to it. Therefore, the maximum temperature at which a gas can be liquefied is just below the critical temperature.

Question 10

The relationship between the vapor pressure of a liquid and temperature can be expressed by the Clausius-Clapeyron equation: Ln[P₂/P₁] = [Δ_vapH/R] × [1/T₁ - 1/T₂] Ethanol (C₂H₅OH)has a normal boiling point of 78.3 °C and,with Δ_vapH = 39.3 kJ/mol.What is the vapor pressure of ethanol at 50.0 °C?

Accepted Answer

First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15. T1 = 78.3 + 273.15 = 351.45 K (normal boiling point) T2 = 50.0 + 273.15 = 323.15 K Next, we can plug in the given values into the Clausius-Clapeyron equation: Ln[P₂/P₁] = [39.3 kJ/mol/R] × [1/351.45 K - 1/323.15 K] We can simplify the equation: Ln[P₂/P₁] = 2.122 Now, we can solve for P₂: P₂/1 atm = e^2.122 P₂ = 8.326 atm Finally, we convert the pressure from atm to Torr: P₂ = 8.326 atm × 760 Torr/atm = 6328.56 Torr To get the vapor pressure at 50.0 °C, we need to use the Antoine equation or look up the vapor pressure of ethanol at 50.0 °C in a table. We can assume that the vapor pressure at 50.0 °C is much less than the atmospheric pressure of 1 atm, so we can subtract the vapor pressure from the atmospheric pressure to get the partial pressure of the ethanol vapor: 1 atm - 6328.56 Torr ≈ 234 Torr Therefore, the answer is B.

Question 11

The normal boiling point of acetone is 56.2 °C and the molar heat of vaporization is 32.0 kJ/mol.At what temperature will acetone boil under a pressure of 50.0 mmHg?

Accepted Answer

The answer of The normal boiling point of acetone is...

Question 12

A 0.90 g sample of liquid water was introduced into an evacuated 2.00 L flask,which was then sealed and heated to 37 °C.What percentage,by mass,of the water remained as liquid? [Vapor pressure of water at 37 °C = 48.2 Torr.]

Accepted Answer

The answer of A 0.90 g sample of liquid water...

Question 13

The factor that has the largest effect on the vapor pressure of a molecular substance is ________.

Accepted Answer

The presence of H-bonding has the largest effect on vapor pressure because it affects the intermolecular forces and the tendency for molecules to escape into the gas phase. H-bonding creates stronger intermolecular forces and therefore a lower vapor pressure.

Question 14

The heat of deposition equals the negative of the heat of ________.

Accepted Answer

The heat of deposition is the negative of the heat of sublimation because deposition is the process where a gas turns directly into a solid, which is the reverse of sublimation (solid to gas).

Question 15

Liquid and vapor phases of a substance become indistinguishable at the ________.

Accepted Answer

The critical point is the temperature and pressure at which the liquid and vapor phases of a substance become indistinguishable, and above which it is not possible to liquify the substance by increasing pressure alone. At this point, the substance has unique properties, including a critical temperature and a critical pressure.

Question 16

Arrange the following compounds in order of increasing viscosity: I.1-propanol,CH₃CH₂CH₂OH; II.1,2-propanediol,CH₃CH(OH)CH₂OH; III.1,2,3-propanetriol, (glycerol),HOCH₂CH(OH)CH₂OH.

Accepted Answer

Viscosity increases with the number of hydroxyl (OH) groups due to increased hydrogen bonding. Therefore, 1-propanol (I) with one OH group has the lowest viscosity, followed by 1,2-propanediol (II) with two OH groups, and 1,2,3-propanetriol (III) with three OH groups has the highest viscosity.

Question 17

Given the data below,determine the normal boiling point of liquid Rb. P₁ = 400 mmHg,t₁ = 620 °C,Δ_vapH = 78.5 kJ/mol

Accepted Answer

The answer of Given the data below,determine the normal boiling...

Question 18

Given the data below,determine the molar enthalpy of vaporization of liquid Rb. Rb,P₁ = 1.00 mmHg,t₁ = 297 °C,P₂ = 400 mmHg,t₂ = 620 °C

Accepted Answer

The molar enthalpy of vaporization can be calculated using the Clausius-Clapeyron equation. The given pressures and temperatures can be used to find the enthalpy change, which is approximately 78.5 kJ/mol for Rb.

Question 19

Given the data below,determine the molar enthalpy of vaporization of COCl₂. COCl₂,P₁ = 40 mmHg,t₁ = -50.3 °C,P₂ = 100 mmHg,t₂ = -35.6 °C

Accepted Answer

The Clausius-Clapeyron equation can be used to calculate the molar enthalpy of vaporization ($\Delta H_{vap}$) using the given data. The equation in its linear form is:$$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$Given:- $P_1 = 40$ mmHg, $T_1 = -50.3$ °C $= 222.85$ K (since $T(K) = T(°C) + 273.15$)- $P_2 = 100$ mmHg, $T_2 = -35.6$ °C $= 237.55$ KFirst, convert pressures to the same units if necessary (in this case, both are in mmHg, so no conversion is needed). Then, solve for $\Delta H_{vap}$:$$\ln\left(\frac{100}{40}\right) = -\frac{\Delta H_{vap}}{8.314}\left(\frac{1}{237.55} - \frac{1}{222.85}\right)$$Solving for $\Delta H_{vap}$ gives a value close to 27.4 kJ/mol, indicating that the correct choice is B.

Question 20

What would be the appropriate equilibrium temperature of a system obtained by adding 25.0 g of ice at 0 °C to 250.0 mL of "hot" (80 °C)coffee,assuming that the heat capacity and density of the coffee are the same as for pure water (4.18 J g^-1 °C^-1 and 0.997 g mL^-1),and also assuming negligible heat transfer with the surroundings? [For water,Δ_fusH = 6.02 kJ/mol.]

Accepted Answer

First, we need to calculate the heat required to melt the ice:
q1 = m × ΔHfus = (25.0 g) × (6.02 kJ/mol) ÷ (18.02 g/mol) = 8.35 kJ
Then, we need to calculate the heat required to raise the temperature of the resulting water from 0 °C to the equilibrium temperature T:
q2 = m × C × ΔT = (275.0 g) × (4.18 J/g°C) × (T - 0 °C)
Since the system is in thermal equilibrium, we know that q1 + q2 = 0:
8.35 kJ + (275.0 g) × (4.18 J/g°C) × (T - 0 °C) = 0
Simplifying and solving for T:
T = 64.9 °C
Rounding to the nearest degree gives us an equilibrium temperature of 65 °C (choice C).

Quiz 12: Intermolecular Forces: Liquids and Solids

Arrange the following compounds in order of increasing boiling point: I.1-propanol,CH₃CH₂CH₂OH; II.1,2-propanediol,CH₃CH(OH) CH₂OH; III.1,2,3-propanetriol, (glycerol) ,HOCH₂CH(OH) CH₂OH.

A substance has a heat of fusion of 61.5 kJ/mol and a heat of deposition of -167.4 kJ/mol.What is the heat of sublimation in kJ/mol?

If a substance has a heat of condensation of -1.46 kJ/g and a heat of sublimation of 4.60 kJ/g,what is its heat of solidification in kJ/g?

If 34 g of a solid with a molar mass = 174 g/mol requires 21.3 kJ to melt it,what is the molar heat of fusion in kJ/mol?

Given the data below,determine the normal boiling point of COCl₂. P₁ = 100 mmHg,t₁ = -35.6 °C,Δ_vapH = 27.4 kJ/mol

How much heat would be released by the condensation of 5.40 g of steam at 100 °C and the subsequent cooling of the water to 25 °C? [Δ_vapH = 40.7 kJ/mol at 100 °C;C_p for H₂O(l) is 4.18 J g^-1 °C^-1]

A liquid has a molar heat of vaporization of 22.7 kJ/mol.Its normal boiling point is 459 K.What is the vapor pressure,in mmHg,at 70 °C?

When the vapor pressure of a liquid equals atmospheric pressure,the temperature of the liquid equals ________.

The maximum temperature at which a gas can be liquefied is just below the ________.

The normal boiling point of acetone is 56.2 °C and the molar heat of vaporization is 32.0 kJ/mol.At what temperature will acetone boil under a pressure of 50.0 mmHg?

A 0.90 g sample of liquid water was introduced into an evacuated 2.00 L flask,which was then sealed and heated to 37 °C.What percentage,by mass,of the water remained as liquid? [Vapor pressure of water at 37 °C = 48.2 Torr.]

The factor that has the largest effect on the vapor pressure of a molecular substance is ________.

The heat of deposition equals the negative of the heat of ________.

Liquid and vapor phases of a substance become indistinguishable at the ________.

Arrange the following compounds in order of increasing viscosity: I.1-propanol,CH₃CH₂CH₂OH; II.1,2-propanediol,CH₃CH(OH) CH₂OH; III.1,2,3-propanetriol, (glycerol) ,HOCH₂CH(OH) CH₂OH.

Given the data below,determine the normal boiling point of liquid Rb. P₁ = 400 mmHg,t₁ = 620 °C,Δ_vapH = 78.5 kJ/mol

Given the data below,determine the molar enthalpy of vaporization of liquid Rb. Rb,P₁ = 1.00 mmHg,t₁ = 297 °C,P₂ = 400 mmHg,t₂ = 620 °C

Given the data below,determine the molar enthalpy of vaporization of COCl₂. COCl₂,P₁ = 40 mmHg,t₁ = -50.3 °C,P₂ = 100 mmHg,t₂ = -35.6 °C

Quiz 12: Intermolecular Forces: Liquids and Solids

Arrange the following compounds in order of increasing boiling point: I.1-propanol,CH3CH2CH2OH; II.1,2-propanediol,CH3CH(OH) CH2OH; III.1,2,3-propanetriol, (glycerol) ,HOCH2CH(OH) CH2OH.

A substance has a heat of fusion of 61.5 kJ/mol and a heat of deposition of -167.4 kJ/mol.What is the heat of sublimation in kJ/mol?

If a substance has a heat of condensation of -1.46 kJ/g and a heat of sublimation of 4.60 kJ/g,what is its heat of solidification in kJ/g?

If 34 g of a solid with a molar mass = 174 g/mol requires 21.3 kJ to melt it,what is the molar heat of fusion in kJ/mol?

Given the data below,determine the normal boiling point of COCl2. P1 = 100 mmHg,t1 = -35.6 °C,ΔvapH = 27.4 kJ/mol

How much heat would be released by the condensation of 5.40 g of steam at 100 °C and the subsequent cooling of the water to 25 °C? [ΔvapH = 40.7 kJ/mol at 100 °C;Cp for H2O(l) is 4.18 J g-1 °C-1]

A liquid has a molar heat of vaporization of 22.7 kJ/mol.Its normal boiling point is 459 K.What is the vapor pressure,in mmHg,at 70 °C?

When the vapor pressure of a liquid equals atmospheric pressure,the temperature of the liquid equals ________.

The maximum temperature at which a gas can be liquefied is just below the ________.

The normal boiling point of acetone is 56.2 °C and the molar heat of vaporization is 32.0 kJ/mol.At what temperature will acetone boil under a pressure of 50.0 mmHg?

A 0.90 g sample of liquid water was introduced into an evacuated 2.00 L flask,which was then sealed and heated to 37 °C.What percentage,by mass,of the water remained as liquid? [Vapor pressure of water at 37 °C = 48.2 Torr.]

The factor that has the largest effect on the vapor pressure of a molecular substance is ________.

The heat of deposition equals the negative of the heat of ________.

Liquid and vapor phases of a substance become indistinguishable at the ________.

Arrange the following compounds in order of increasing viscosity: I.1-propanol,CH3CH2CH2OH; II.1,2-propanediol,CH3CH(OH) CH2OH; III.1,2,3-propanetriol, (glycerol) ,HOCH2CH(OH) CH2OH.

Given the data below,determine the normal boiling point of liquid Rb. P1 = 400 mmHg,t1 = 620 °C,ΔvapH = 78.5 kJ/mol

Given the data below,determine the molar enthalpy of vaporization of liquid Rb. Rb,P1 = 1.00 mmHg,t1 = 297 °C,P2 = 400 mmHg,t2 = 620 °C

Given the data below,determine the molar enthalpy of vaporization of COCl2. COCl2,P1 = 40 mmHg,t1 = -50.3 °C,P2 = 100 mmHg,t2 = -35.6 °C

Arrange the following compounds in order of increasing boiling point: I.1-propanol,CH₃CH₂CH₂OH; II.1,2-propanediol,CH₃CH(OH) CH₂OH; III.1,2,3-propanetriol, (glycerol) ,HOCH₂CH(OH) CH₂OH.

Given the data below,determine the normal boiling point of COCl₂. P₁ = 100 mmHg,t₁ = -35.6 °C,Δ_vapH = 27.4 kJ/mol

How much heat would be released by the condensation of 5.40 g of steam at 100 °C and the subsequent cooling of the water to 25 °C? [Δ_vapH = 40.7 kJ/mol at 100 °C;C_p for H₂O(l) is 4.18 J g^-1 °C^-1]

Arrange the following compounds in order of increasing viscosity: I.1-propanol,CH₃CH₂CH₂OH; II.1,2-propanediol,CH₃CH(OH) CH₂OH; III.1,2,3-propanetriol, (glycerol) ,HOCH₂CH(OH) CH₂OH.

Given the data below,determine the normal boiling point of liquid Rb. P₁ = 400 mmHg,t₁ = 620 °C,Δ_vapH = 78.5 kJ/mol

Given the data below,determine the molar enthalpy of vaporization of liquid Rb. Rb,P₁ = 1.00 mmHg,t₁ = 297 °C,P₂ = 400 mmHg,t₂ = 620 °C

Given the data below,determine the molar enthalpy of vaporization of COCl₂. COCl₂,P₁ = 40 mmHg,t₁ = -50.3 °C,P₂ = 100 mmHg,t₂ = -35.6 °C