Question 1

Freon-113,C₂Cl₃F₃,has an enthalpy of vaporization of 27.0 kJ/mol and a normal boiling point of 48.0 °C.What is the vapor pressure (in atm)of Freon-113 at 22.0 °C? (R = 8.314 J/K⋅mol)

Accepted Answer

Using the Clausius-Clapeyron equation, we have:

ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)

where P2 is the vapor pressure at the boiling point (1 atm), P1 is the vapor pressure at the given temperature (unknown), ΔHvap is the enthalpy of vaporization, R is the gas constant, T1 is the boiling point temperature, and T2 is the given temperature.

Substituting the given values, we have:

ln(1/P1) = (27.0*10^3 J/mol) / (8.314 J/K⋅mol) * (1/321 K - 1/295 K)

Solving for ln(1/P1), we get:

ln(1/P1) = -10.471

Taking the exponential of both sides, we get:

1/P1 = e^-10.471

Solving for P1, we get:

P1 = 0.41 atm

Therefore, the vapor pressure of Freon-113 at 22.0°C is 0.41 atm, which is answer C.

Question 2

Xenon has an enthalpy of vaporization of 12.6 kJ/mol and a vapor pressure of 1.00 atm at -108.0 °C.What is the vapor pressure of xenon at -148.0 °C? (R = 8.314 J/K⋅mol)

Accepted Answer

To find the vapor pressure at a new temperature, we use the Clausius-Clapeyron equation: $\ln(\frac{P_2}{P_1}) = \frac{-\Delta H_{vap}}{R}(\frac{1}{T_2} - \frac{1}{T_1})$, where $P_1$ and $P_2$ are the initial and final pressures, $T_1$ and $T_2$ are the initial and final temperatures in Kelvin, $\Delta H_{vap}$ is the enthalpy of vaporization, and $R$ is the gas constant. Convert temperatures from Celsius to Kelvin: $T_1 = -108.0 °C = 165 K$ and $T_2 = -148.0 °C = 125 K$. Solve for $P_2$ to find the vapor pressure at -148.0 °C.

Question 3

For a particular liquid,raising its temperature from 331 K to 351 K causes its vapor pressure to double.What is the enthalpy of vaporization of this liquid? (R = 8.314 J/K · mol)

Accepted Answer

The enthalpy of vaporization can be calculated using the Clausius-Clapeyron equation: $\Delta H_{vap} = \frac{R \cdot \Delta T \cdot \ln(\frac{P_2}{P_1})}{\frac{1}{T_1} - \frac{1}{T_2}}$, where $P_2/P_1 = 2$ (vapor pressure doubles), $T_1 = 331 K$, $T_2 = 351 K$, and $R = 8.314 J/K \cdot mol$. Plugging these values in, we find $\Delta H_{vap} \approx 34 kJ/mol$.

Question 4

The best explanation for the existence of a meniscus observed when water is placed in a glass tube of small diameter is

Accepted Answer

The meniscus occurs because the attractive forces between the water molecules and the walls of the container are greater than the attractive forces between the water molecules themselves. This causes the water to "climb up" the walls of the container, creating a curved surface.

Question 5

Capillary action is ____

Accepted Answer

Capillary action refers to the drawing of a liquid up the inside of a small-bore tube due to the adhesive forces between the liquid and the tube being greater than the cohesive forces between the liquid molecules. This phenomenon is responsible for many important processes in nature, such as the movement of water through plants and the rise of blood in capillaries.

Question 6

A liquid has an enthalpy of vaporization of 30.3 kJ/mol.At 262 K it has a vapor pressure of 121 mmHg.What is the normal boiling point of this liquid? (R = 8.314 J/(K· mol))

Accepted Answer

We can use the Clausius-Clapeyron equation:

ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)

where P1 and T1 are the given values, P2 is the pressure at the normal boiling point (1 atm), T2 is the normal boiling point (what we are solving for), ΔHvap is the enthalpy of vaporization, and R is the gas constant.

Plugging in the values we get:

ln(760/121) = (-30300 J/mol / (8.314 J/(K·mol))) * (1/T2 - 1/262K)

Simplifying:

ln(6.28) = (-3642.7) * (1/T2 - 0.0038)

ln(6.28)/(-3642.7) = 1/T2 - 0.0038

1/T2 = ln(6.28)/(-3642.7) + 0.0038

T2 = 302 K

Therefore, the answer is B: 302 K.

Question 7

Ethanol has an enthalpy of vaporization of 42.3 kJ/mol.The compound has a vapor pressure of 1.00 atm at 78.3 °C.At what temperature is the vapor pressure equal to 0.800 atm? (R = 8.314 J/K ⋅ mol)

Accepted Answer

First, we need to use the Clausius-Clapeyron equation to determine the temperature at which the vapor pressure is 0.800 atm:
ln(P₂/P₁) = -(ΔH_vap/R) × (1/T₂ - 1/T₁)
Since we know P₁ (1.00 atm) and T₁ (78.3°C + 273.15 = 351.45 K) and P₂ (0.800 atm), we can solve for T₂:
ln(0.800/1.00) = -(42.3 × 10³/R) × (1/T₂ - 1/351.45)
-0.2231 = -(42.3 × 10³/8.314) × (1/T₂ - 1/351.45)
-0.2231 = -5095.5 × (1/T₂ - 0.002839)
1/T₂ = 0.0135 + 1/351.45
1/T₂ = 0.016939
T₂ = 59.0°C + 273.15 = 332.15 K
Therefore, the temperature at which the vapor pressure of ethanol is 0.800 atm is 59.0°C, which is closest to answer choice D: 73.0°C.

Question 8

Which of the following is the slope of the line that illustrates the relationship between the natural logarithm of the vapor pressure of a gas and the reciprocal of its temperature (in kelvin)?

Accepted Answer

The answer of Which of the following is the slope...

Question 9

If 4.61 g of water is sealed in an evacuated 3.35 L flask and heated to the normal boiling point of 373 K,what is the pressure in the flask? (R = 0.08206 L⋅atm/mol⋅K)

Accepted Answer

The pressure in the flask can be calculated using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. First, calculate the number of moles of water (H2O) vapor: mass (m) = 4.61 g, molar mass of H2O = 18.015 g/mol, so n = m/M = 4.61 g / 18.015 g/mol = 0.256 mol. Then, using the ideal gas law: P = nRT/V = (0.256 mol)(0.08206 L⋅atm/mol⋅K)(373 K) / 3.35 L = 1.00 atm.

Question 10

Normal boiling point is defined as

Accepted Answer

Normal boiling point is defined as the temperature at which the vapor pressure of a liquid equals 1 atm.

Question 11

What is responsible for capillary action,a property of liquids?

Accepted Answer

The answer of What is responsible for capillary action,a property...

Question 12

The vapor pressure of nitric acid is 26.6 mmHg at 10.0 °C and 208 mmHg at 50.0 °C.Using this information,calculate the heat of vaporization (Δ_vapH)of nitric acid.(R = 8.314 J/K ⋅ mol)

Accepted Answer

The Clausius-Clapeyron equation can be used to calculate the heat of vaporization ($\Delta H_{vap}$) from the vapor pressure at two different temperatures. The equation in its logarithmic 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:- $P_1 = 26.6$ mmHg at $T_1 = 10.0$ °C = 283.15 K- $P_2 = 208$ mmHg at $T_2 = 50.0$ °C = 323.15 K- $R = 8.314$ J/K ⋅ molFirst, convert pressures to the same units as R (if needed, but here we can cancel units since both pressures are in the same unit and we are interested in their ratio).$$\ln\left(\frac{208}{26.6}\right) = -\frac{\Delta H_{vap}}{8.314}\left(\frac{1}{323.15} - \frac{1}{283.15}\right)$$Solving for $\Delta H_{vap}$:$$\Delta H_{vap} = -\ln\left(\frac{208}{26.6}\right) \times \frac{8.314}{\left(\frac{1}{323.15} - \frac{1}{283.15}\right)}$$$$\Delta H_{vap} \approx 39.1 \, \text{kJ/mol}$$Therefore, the heat of vaporization of nitric acid is approximately 39.1 kJ/mol.

Question 13

Which of the following is not a characteristic of volatile liquids?

Accepted Answer

Volatile liquids have weak cohesive forces, which allow them to easily evaporate into vapor. Therefore, option C, which states that volatile liquids have strong cohesive forces, is not a characteristic of volatile liquids.

Question 14

In a certain mountain range,water boils at 93°C.What is the atmospheric pressure under these conditions? The enthalpy of vaporization of water at 100°C is 40.7 kJ/mol.(R = 8.314 J/K·mol; 1 atm = 760 mmHg)

Accepted Answer

Using the Clausius-Clapeyron equation, we can calculate the atmospheric pressure at which water boils at 93°C. The equation is: ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1). By substituting the given values and solving for P2, we find that the pressure is approximately 591 mmHg.

Question 15

At 80.0 °C,water has an equilibrium vapor pressure of 355.1 mm Hg.If 4.22 g H₂O is sealed in an evacuated 5.00 L flask and heated to 80.0 °C,what mass of H₂O will be found in the gas phase when liquid-vapor equilibrium is established? Assume any liquid remaining in the flask has a negligible volume.(R = 0.08206 L⋅atm/mol⋅K,1 atm = 760 mm Hg)

Accepted Answer

The answer of At 80.0 °C,water has an equilibrium vapor...

Question 16

A particular compound has an enthalpy of vaporization of 29600 J/mol.At 274 K it has a vapor pressure of 103 mmHg.What is its vapor pressure at 309 K? (R = 8.31 J/K·mol; 1 atm = 760 mmHg)

Accepted Answer

The answer of A particular compound has an enthalpy of...

Question 17

Mount Everest rises to a height of 8.850 × 10³ m above sea level.At this height,the atmospheric pressure is 231 mm Hg.At what temperature (in °C)will water boil at the summit of Mount Everest? The vapor pressure of water at 373 K is 760.0 mm Hg.(Δ_vapH° for H₂O = 40.7 kJ/mol and R = 8.314 J/K ⋅ mol)

Accepted Answer

First, we need to calculate the atmospheric pressure at the summit of Mount Everest using the equation:

P = P₀e^(-Mgh/RT)

Where P₀ is the atmospheric pressure at sea level (760 mm Hg), M is the molar mass of air (0.02897 kg/mol), g is the acceleration due to gravity (9.81 m/s²), h is the height above sea level (8.850 × 10^3 m), R is the gas constant (8.314 J/K⋅mol), and T is the temperature in Kelvin.

Plugging in the values, we get:

P = (760 mm Hg) × exp[-(0.02897 kg/mol) × (9.81 m/s²) × (8.850 × 10^3 m) / (8.314 J/K⋅mol × 288.15 K)]

P = 231 mm Hg

Now we can use the Clausius-Clapeyron equation to determine the boiling point of water at this pressure:

ln(P₀/P) = ΔH_vap/R × (1/T - 1/T₀)

Where P₀ is the vapor pressure of water at 373 K (760 mm Hg), T₀ is 373 K, ΔH_vap is the enthalpy of vaporization of water (40.7 kJ/mol), and R is the gas constant (8.314 J/K⋅mol).

Solving for T, we get:

T = ΔH_vap/R × (1/ln(P₀/P) + 1/T₀)^-1

Plugging in the values, we get:

T = (40.7 kJ/mol)/(8.314 J/K⋅mol) × (1/ln(760 mm Hg/231 mm Hg) + 1/373 K)^-1

T = 69.0 °C

Therefore, the boiling point of water at the summit of Mount Everest is 69.0 °C, which is choice B.

Question 18

Sulfur dioxide has a vapor pressure of 462.7 mm Hg at -21.0 °C and a vapor pressure of 140.5 mm Hg at -44.0 °C.What is the enthalpy of vaporization of sulfur dioxide? (R = 8.314 J/K⋅mol)

Accepted Answer

We can use the Clausius-Clapeyron Equation:

ln(P₂/P₁) = -(ΔHvap/R) * (1/T₂ - 1/T₁)

where P₁ and P₂ are the vapor pressures at temperatures T₁ and T₂ respectively.

Rearranging and plugging in the values, we get:

ΔHvap = -(R * ln(P₂/P₁)) / (1/T₂ - 1/T₁)

ΔHvap = -((8.314 J/K⋅mol) * ln(140.5/462.7 mm Hg)) / (1/(-44.0 + 273.15) K - 1/(-21.0 + 273.15) K)

ΔHvap = 24.9 kJ/mol

Therefore, the enthalpy of vaporization of sulfur dioxide is 24.9 kJ/mol, which is closest to choice D.

Question 19

The heat of vaporization of benzene (C₆H₆)is 30.7 kJ/mol at its boiling point of 80.1 °C.How much energy in the form of heat is required to vaporize 110 g benzene at its boiling point?

Accepted Answer

The answer of The heat of vaporization of benzene (C₆H₆)is...

Question 20

A cup of oil takes longer to pour from a glass than a cup of water.A liquid's resistance to flow is referred to as its ________.

Accepted Answer

The answer of A cup of oil takes longer to...

Quiz 12: Intermolecular Forces and Liquids

Freon-113,C₂Cl₃F₃,has an enthalpy of vaporization of 27.0 kJ/mol and a normal boiling point of 48.0 °C.What is the vapor pressure (in atm) of Freon-113 at 22.0 °C? (R = 8.314 J/K⋅mol)

Xenon has an enthalpy of vaporization of 12.6 kJ/mol and a vapor pressure of 1.00 atm at -108.0 °C.What is the vapor pressure of xenon at -148.0 °C? (R = 8.314 J/K⋅mol)

For a particular liquid,raising its temperature from 331 K to 351 K causes its vapor pressure to double.What is the enthalpy of vaporization of this liquid? (R = 8.314 J/K · mol)

The best explanation for the existence of a meniscus observed when water is placed in a glass tube of small diameter is

Capillary action is ____

A liquid has an enthalpy of vaporization of 30.3 kJ/mol.At 262 K it has a vapor pressure of 121 mmHg.What is the normal boiling point of this liquid? (R = 8.314 J/(K· mol) )

Ethanol has an enthalpy of vaporization of 42.3 kJ/mol.The compound has a vapor pressure of 1.00 atm at 78.3 °C.At what temperature is the vapor pressure equal to 0.800 atm? (R = 8.314 J/K ⋅ mol)

Which of the following is the slope of the line that illustrates the relationship between the natural logarithm of the vapor pressure of a gas and the reciprocal of its temperature (in kelvin) ?

If 4.61 g of water is sealed in an evacuated 3.35 L flask and heated to the normal boiling point of 373 K,what is the pressure in the flask? (R = 0.08206 L⋅atm/mol⋅K)

Normal boiling point is defined as

What is responsible for capillary action,a property of liquids?

The vapor pressure of nitric acid is 26.6 mmHg at 10.0 °C and 208 mmHg at 50.0 °C.Using this information,calculate the heat of vaporization (Δ_vapH) of nitric acid.(R = 8.314 J/K ⋅ mol)

Which of the following is not a characteristic of volatile liquids?

In a certain mountain range,water boils at 93°C.What is the atmospheric pressure under these conditions? The enthalpy of vaporization of water at 100°C is 40.7 kJ/mol.(R = 8.314 J/K·mol; 1 atm = 760 mmHg)

A particular compound has an enthalpy of vaporization of 29600 J/mol.At 274 K it has a vapor pressure of 103 mmHg.What is its vapor pressure at 309 K? (R = 8.31 J/K·mol; 1 atm = 760 mmHg)

Sulfur dioxide has a vapor pressure of 462.7 mm Hg at -21.0 °C and a vapor pressure of 140.5 mm Hg at -44.0 °C.What is the enthalpy of vaporization of sulfur dioxide? (R = 8.314 J/K⋅mol)

The heat of vaporization of benzene (C₆H₆) is 30.7 kJ/mol at its boiling point of 80.1 °C.How much energy in the form of heat is required to vaporize 110 g benzene at its boiling point?

A cup of oil takes longer to pour from a glass than a cup of water.A liquid's resistance to flow is referred to as its ________.

Quiz 12: Intermolecular Forces and Liquids

Freon-113,C2Cl3F3,has an enthalpy of vaporization of 27.0 kJ/mol and a normal boiling point of 48.0 °C.What is the vapor pressure (in atm) of Freon-113 at 22.0 °C? (R = 8.314 J/K⋅mol)

Xenon has an enthalpy of vaporization of 12.6 kJ/mol and a vapor pressure of 1.00 atm at -108.0 °C.What is the vapor pressure of xenon at -148.0 °C? (R = 8.314 J/K⋅mol)

For a particular liquid,raising its temperature from 331 K to 351 K causes its vapor pressure to double.What is the enthalpy of vaporization of this liquid? (R = 8.314 J/K · mol)

The best explanation for the existence of a meniscus observed when water is placed in a glass tube of small diameter is

Capillary action is ____

A liquid has an enthalpy of vaporization of 30.3 kJ/mol.At 262 K it has a vapor pressure of 121 mmHg.What is the normal boiling point of this liquid? (R = 8.314 J/(K· mol) )

Ethanol has an enthalpy of vaporization of 42.3 kJ/mol.The compound has a vapor pressure of 1.00 atm at 78.3 °C.At what temperature is the vapor pressure equal to 0.800 atm? (R = 8.314 J/K ⋅ mol)

Which of the following is the slope of the line that illustrates the relationship between the natural logarithm of the vapor pressure of a gas and the reciprocal of its temperature (in kelvin) ?

If 4.61 g of water is sealed in an evacuated 3.35 L flask and heated to the normal boiling point of 373 K,what is the pressure in the flask? (R = 0.08206 L⋅atm/mol⋅K)

Normal boiling point is defined as

What is responsible for capillary action,a property of liquids?

The vapor pressure of nitric acid is 26.6 mmHg at 10.0 °C and 208 mmHg at 50.0 °C.Using this information,calculate the heat of vaporization (ΔvapH) of nitric acid.(R = 8.314 J/K ⋅ mol)

Which of the following is not a characteristic of volatile liquids?

In a certain mountain range,water boils at 93°C.What is the atmospheric pressure under these conditions? The enthalpy of vaporization of water at 100°C is 40.7 kJ/mol.(R = 8.314 J/K·mol; 1 atm = 760 mmHg)

A particular compound has an enthalpy of vaporization of 29600 J/mol.At 274 K it has a vapor pressure of 103 mmHg.What is its vapor pressure at 309 K? (R = 8.31 J/K·mol; 1 atm = 760 mmHg)

Sulfur dioxide has a vapor pressure of 462.7 mm Hg at -21.0 °C and a vapor pressure of 140.5 mm Hg at -44.0 °C.What is the enthalpy of vaporization of sulfur dioxide? (R = 8.314 J/K⋅mol)

The heat of vaporization of benzene (C6H6) is 30.7 kJ/mol at its boiling point of 80.1 °C.How much energy in the form of heat is required to vaporize 110 g benzene at its boiling point?

A cup of oil takes longer to pour from a glass than a cup of water.A liquid's resistance to flow is referred to as its ________.

Freon-113,C₂Cl₃F₃,has an enthalpy of vaporization of 27.0 kJ/mol and a normal boiling point of 48.0 °C.What is the vapor pressure (in atm) of Freon-113 at 22.0 °C? (R = 8.314 J/K⋅mol)

The vapor pressure of nitric acid is 26.6 mmHg at 10.0 °C and 208 mmHg at 50.0 °C.Using this information,calculate the heat of vaporization (Δ_vapH) of nitric acid.(R = 8.314 J/K ⋅ mol)

The heat of vaporization of benzene (C₆H₆) is 30.7 kJ/mol at its boiling point of 80.1 °C.How much energy in the form of heat is required to vaporize 110 g benzene at its boiling point?