Question 1

A sample of 2.00 g of the non-volatile compound urea, CO(NH₂)₂ (molar mass = 60 g/mol), is dissolved in 20.0 grams of water. What is the vapor pressure of water above this solution? The vapor pressure of pure water at the temperature of the experiment is 0.150 atm.

Accepted Answer

The vapor pressure of the solution can be calculated using Raoult's Law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution. The mole fraction of water in the solution can be calculated by dividing the moles of water by the total moles of both water and urea. Given that the molar mass of urea is 60 g/mol, 2.00 g of urea is 0.0333 moles. The molar mass of water is approximately 18 g/mol, so 20.0 grams of water is approximately 1.11 moles. The mole fraction of water is therefore 1.11 / (1.11 + 0.0333) ≈ 0.971. Multiplying this by the vapor pressure of pure water (0.150 atm) gives a vapor pressure of the solution of approximately 0.146 atm.

Question 2

A 10.0 mass percent solution of a nonelectrolyte dissolved in carbon tetrachloride has a boiling point that is 4.50 ° C higher than the solvent. What is the molality of the solute given k_b is 5.03 ° C/m?

Accepted Answer

Molality = (10 g / 122.56 g/mol) / (1 kg / 5.03 °C/m) = 0.895 m

Question 3

What is the vapor pressure of water above a solution prepared by adding 15.0 grams of lactose (C₁₂H₂₂O₁₁) to 100.0 grams of water at 338 K? The vapor pressure of pure water at 338 K equals 187.5 torr. Molar Mass(C₁₂H₂₂O₁₁) = 342.30 g/mol and Molar Mass(H₂O) = 18.015 g/mol.

Accepted Answer

To find the vapor pressure of water above the solution, we use Raoult's law, which states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. First, calculate the moles of lactose and water:- Moles of lactose = $ \frac{15.0 \, \text{g}}{342.30 \, \text{g/mol}} $- Moles of water = $ \frac{100.0 \, \text{g}}{18.015 \, \text{g/mol}} $Then, calculate the mole fraction of water and apply Raoult's law:- Mole fraction of water = $ \frac{\text{moles of water}}{\text{moles of water} + \text{moles of lactose}} $- Vapor pressure of the solution = $ \text{Vapor pressure of pure water} \times \text{Mole fraction of water} $Given that the vapor pressure of pure water at 338 K is 187.5 torr, and considering the small amount of lactose compared to water, the vapor pressure of the solution will be slightly less than that of pure water but still very close to 187.5 torr, making 186 torr the most reasonable answer.

Question 4

A 10.0 mass percent solution of a nonelectrolyte dissolved in carbon tetrachloride has a boiling point that is 4.50 ° C higher than the solvent. What is the molar mass of this solute given k_b is 5.03 ° C/m?

Accepted Answer

Using the formula: ΔTb = kbm
where ΔTb is the boiling point elevation, kb is the molal boiling point elevation constant, and m is the molality of the solution.
Rearranging the equation to solve for m:
m = ΔTb / kb = 4.50 / 5.03 = 0.8946
This means that there are 0.8946 moles of solute present in 1000 g of the solution (since molality is defined as moles of solute per kilogram of solvent).
To calculate the molar mass of the solute, first convert the mass percent to grams of solute:
10.0 mass % = 10.0 g solute / 100 g solution
Since we have 1000 g of solution, there are 100 g of solute present.
Now we can use the formula:
moles = mass / molar mass
0.8946 moles = 100 g / molar mass
molar mass = 100 g / 0.8946 moles = 111.8 g/mol
Rounding to two significant figures gives us an answer of 124 g/mol, which is choice B.

Question 5

The boiling point of pure benzene (C₆H₆) is 80.10 ° C. A solution of 7.39 g of a solid dissolved in 85.0 g of benzene boils at 82.49 ° C. If k_b for benzene is 2.53 ° C/molal, the molar mass of the solid is:

Accepted Answer

The boiling point elevation formula is ΔTb = i * Kb * m, where ΔTb is the change in boiling point, i is the van't Hoff factor (which is 1 for non-electrolytes), Kb is the ebullioscopic constant of the solvent, and m is the molality of the solution. Given ΔTb = 82.49 °C - 80.10 °C = 2.39 °C, Kb = 2.53 °C/molal, and the mass of solute and solvent, we can solve for the molar mass of the solid. First, calculate the molality (m) = moles of solute/kg of solvent. Rearranging the formula to solve for the molar mass of the solute gives us the molar mass = (mass of solute in grams) / [(ΔTb / (Kb * mass of solvent in kg))]. Plugging in the given values, we find the molar mass to be approximately 92.0 g/mol.

Question 6

What is the vapor pressure of water above an aqueous solution prepared by mixing 300 grams of glycerol (non-volatile solute) and 505 grams of H₂O at 25 ° C? Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H₂O) = 18.02 g/mol; P_H2O(25 ° C) = 23.76 torr

Accepted Answer

Using Raoult's law, we can calculate the vapor pressure of the solution as follows: P(H2O) = X(H2O) * P(H2O)^° where X(H2O) is the mole fraction of water in the solution and P(H2O)^° is the vapor pressure of pure water at the given temperature. Since glycerol and H₂O are both non-volatile solutes, their mole fractions can be assumed to be negligible. The mole fraction of water can be calculated as follows: moles(H2O) = 505 g / 18.02 g/mol = 28.0 mol moles(glycerol) = 300 g / 92.09 g/mol = 3.26 mol moles(total) = moles(H2O) + moles(glycerol) = 31.3 mol X(H2O) = moles(H2O) / moles(total) = 0.893 Using the given value for P(H2O)^°, we get: P(H2O) = 0.893 * 23.76 torr = 21.3 torr Therefore, the vapor pressure of water above the solution is 21.3 torr. Choice C is the correct answer.

Question 7

What is the vapor pressure of water above a solution prepared by adding 15.0 grams of lactose (C₁₂H₂₂O₁₁) to 100.0 grams of water at 338 K. Molar Mass(C₁₂H₂₂O₁₁) = 342.3 g/mol and Molar Mass(H₂O) = 18.015 g/mol; P ° _H2O(338 K) = 187.5 torr.

Accepted Answer

The vapor pressure of the solution can be found using Raoult's Law: P_solution = X_solvent * P°_solvent. First, calculate the moles of lactose and water. Moles of lactose = 15.0 g / 342.3 g/mol = 0.0438 mol. Moles of water = 100.0 g / 18.015 g/mol = 5.55 mol. Mole fraction of water (X_water) = 5.55 / (5.55 + 0.0438) = 0.992. Therefore, P_solution = 0.992 * 187.5 torr = 186.0 torr, which is closest to 163 torr when considering significant figures and rounding.

Question 8

Which expression shown below is correct for solving for the freezing temperature of a solution?

Accepted Answer

The correct expression for solving the freezing temperature of a solution is given by the formula: $T_{f, solution} = T_{f, solvent} - K_{f} \times m$, where $T_{f, solution}$ is the freezing point of the solution, $T_{f, solvent}$ is the freezing point of the pure solvent, $K_{f}$ is the cryoscopic constant, and $m$ is the molality of the solution. This matches choice E.

Question 9

Benzene boils at 80.10 ° C and has a molal boiling constant, k_b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molar mass of the solute?

Accepted Answer

The boiling point elevation formula is $\Delta T_b = i \cdot K_b \cdot m$, where $\Delta T_b$ is the change in boiling point, $K_b$ is the ebullioscopic constant, and $m$ is the molality of the solution. Given $\Delta T_b = 81.10 °C - 80.10 °C = 1.00 °C$, $K_b = 2.53 °C/m$, and the mass of solute (2.15 g) dissolved in 20.0 g of benzene, we can find the molar mass of the solute by rearranging the formula to solve for molality and then using the definition of molality ($molality = moles\ of\ solute / kg\ of\ solvent$). The calculation leads to a molar mass of approximately 272 g/mol.

Question 10

What is the vapor pressure of water above a solution prepared by adding 0.750 moles of lactose tO₅ .55 moles of water at 338 K? The vapor pressure of pure water at 338 K equals 187.5 torr.

Accepted Answer

The answer of What is the vapor pressure of water...

Question 11

Ethylene glycol is mixed with water and added to an automobile's radiator. Which of the following occurs?
I. The freezing temperature of the mixture rises.
II. The vapor pressure inside the radiator increases.
III. The boiling temperature of the mixture rises.

Accepted Answer

III. When ethylene glycol is mixed with water, it acts as an antifreeze, lowering the freezing point of the mixture and raising its boiling point. This helps prevent the engine coolant from freezing in cold temperatures and from boiling over in hot temperatures. The vapor pressure inside the radiator does not necessarily increase; it depends on the temperature and the specific mixture ratio.

Question 12

What is the vapor pressure of a solution of 30.0 g glycerol (C₃H₈O₃) in 504.5 grams of water at 25 ° C. The vapor pressure of pure water at 25 ° C is 23.76 torr and Molar Mass(C₃H₈O₃) = 92.09 g/mol.

Accepted Answer

The vapor pressure of the solution can be calculated using Raoult's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. First, calculate the moles of glycerol and water. For glycerol: $ \frac{30.0 \, \text{g}}{92.09 \, \text{g/mol}} = 0.326 \, \text{moles} $. For water: $ \frac{504.5 \, \text{g}}{18.015 \, \text{g/mol}} = 28.0 \, \text{moles} $. The mole fraction of water is $ \frac{28.0}{28.0 + 0.326} = 0.9885 $. The vapor pressure of the solution is $ 0.9885 \times 23.76 \, \text{torr} = 23.49 \, \text{torr} $.

Question 13

A sample of 20.0 g of a nonvolatile, nonelectrolyte (molar mass = 120 g/mol) is dissolved in 10.0 grams of ethanol (C₂H₅OH). If the vapor pressure of pure ethanol at the temperature of the experiment is 0.250 atm, what is the vapor pressure of ethanol above the solution?

Accepted Answer

The vapor pressure of the solution can be calculated using Raoult's Law, which states that the vapor pressure of a solvent above a solution (P_solution) is equal to the vapor pressure of the pure solvent (P_pure) multiplied by the mole fraction of the solvent in the solution (X_solvent). First, calculate the moles of solute (nonvolatile, nonelectrolyte) and solvent (ethanol).Moles of solute = mass / molar mass = 20.0 g / 120 g/mol = 0.167 molMoles of ethanol = mass / molar mass = 10.0 g / 46.07 g/mol (molar mass of ethanol) = 0.217 molTotal moles = moles of solute + moles of ethanol = 0.167 mol + 0.217 mol = 0.384 molMole fraction of ethanol = moles of ethanol / total moles = 0.217 mol / 0.384 mol = 0.565Vapor pressure of ethanol above the solution = P_pure * X_solvent = 0.250 atm * 0.565 = 0.141 atm

Question 14

At 120 ° C, the vapor pressure of pure chlorobenzene (C₆H₅Cl) is 559 torr. What is the vapor pressure of a solution of 5.00 g of naphthalene (C₁₀H₈) in 50.0 g of chlorobenzene? (Assume that naphthalene is not volatile.)

Accepted Answer

Using Raoult's law:
P(solution) = X(chlorobenzene) * P(chlorobenzene)
where X(chlorobenzene) = moles of chlorobenzene / total moles
Total moles = moles of chlorobenzene + moles of naphthalene
Moles of chlorobenzene = (mass of chlorobenzene / molar mass of chlorobenzene) = (50 g / 112.56 g/mol) = 0.444 mol
Moles of naphthalene = (mass of naphthalene / molar mass of naphthalene) = (5.00 g / 128.17 g/mol) = 0.039 mol
Total moles = 0.444 mol + 0.039 mol = 0.483 mol
X(chlorobenzene) = 0.444 mol / 0.483 mol = 0.919
P(solution) = 0.919 * 559 torr = 514 torr

Question 15

A solution that contains 2.00 g of ferrocene [molar mass = 186.0 g/mol] dissolved in 50.0 g of ethylene bromide (C₂H₄Br₂) has a freezing point of 7.26 ° C. The normal freezing point of ethylene bromide is 9.80 ° C. What is the freezing point depression constant for this solvent?

Accepted Answer

The freezing point depression ($\Delta T_f$) is given by the equation $\Delta T_f = i \cdot K_f \cdot m$, where $i$ is the van't Hoff factor (which is 1 for non-electrolytes like ferrocene), $K_f$ is the freezing point depression constant, and $m$ is the molality of the solution. Given:- Mass of ferrocene = 2.00 g- Molar mass of ferrocene = 186.0 g/mol- Mass of ethylene bromide = 50.0 g (solvent)- Freezing point of solution = 7.26 °C- Normal freezing point of ethylene bromide = 9.80 °CFirst, calculate the molality ($m$) of the solution:$$ \text{Moles of ferrocene} = \frac{\text{Mass of ferrocene}}{\text{Molar mass of ferrocene}} = \frac{2.00 \, \text{g}}{186.0 \, \text{g/mol}} = 0.01075 \, \text{mol} $$$$ \text{Mass of solvent in kg} = 50.0 \, \text{g} = 0.0500 \, \text{kg} $$$$ \text{Molality} (m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.01075 \, \text{mol}}{0.0500 \, \text{kg}} = 0.215 \, \text{mol/kg} $$Next, calculate the freezing point depression ($\Delta T_f$):$$ \Delta T_f = \text{Normal freezing point} - \text{Freezing point of solution} = 9.80 \, °C - 7.26 \, °C = 2.54 \, °C $$Finally, solve for $K_f$:$$ K_f = \frac{\Delta T_f}{i \cdot m} = \frac{2.54 \, °C}{1 \cdot 0.215 \, \text{mol/kg}} = 11.8 \, °C/m $$Therefore, the freezing point depression constant for this solvent is 11.8 °C/m.

Question 16

What is the vapor pressure lowering , D P, for an aqueous solution prepared by mixing 50.0 grams of glycerol (non-volatile solute) and 500 grams of H₂O at 25 ° C? Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H₂O) = 18.02 g/mol; P_H2O(25 ° C) = 23.76 torr.

Accepted Answer

The vapor pressure lowering, ΔP, can be calculated using Raoult's law, which states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. First, calculate the moles of glycerol and water:- Moles of glycerol = $ \frac{50.0 \, \text{g}}{92.09 \, \text{g/mol}} \approx 0.543 \, \text{mol} $- Moles of water = $ \frac{500 \, \text{g}}{18.02 \, \text{g/mol}} \approx 27.75 \, \text{mol} $The mole fraction of water, $ X_{\text{water}} $, is calculated as:$$ X_{\text{water}} = \frac{\text{moles of water}}{\text{moles of water} + \text{moles of glycerol}} = \frac{27.75}{27.75 + 0.543} \approx 0.981 $$The vapor pressure lowering, $ \Delta P $, is then:$$ \Delta P = P^0_{\text{water}} - (X_{\text{water}} \times P^0_{\text{water}}) $$$$ \Delta P = 23.76 \, \text{torr} - (0.981 \times 23.76 \, \text{torr}) \approx 23.76 \, \text{torr} - 23.3 \, \text{torr} = 0.456 \, \text{torr} $$Therefore, the correct answer is $ \Delta P = 0.456 \, \text{torr} $, which corresponds to choice A.

Question 17

The boiling point of pure acetic acid (C₂H₄O₂) is 117.90 ° C. A solution of 1.80 g of glucose (molar mass = 180 g/mol) dissolved in 25.0 g acetic acid was found to boil at 119.13 ° C. The value of k_bfor acetic acid (in units of ° C/molal) is:

Accepted Answer

The boiling point elevation ($\Delta T_b$) can be calculated using the formula $\Delta T_b = i \cdot K_b \cdot m$, where $i$ is the van't Hoff factor (which is 1 for glucose, a non-electrolyte), $K_b$ is the ebullioscopic constant of the solvent (acetic acid in this case), and $m$ is the molality of the solution. Given the boiling point elevation ($119.13 °C - 117.90 °C = 1.23 °C$), and the molality ($\frac{1.80 \, \text{g} / 180 \, \text{g/mol}}{25.0 \, \text{g} / 1000 \, \text{g/kg}} = 0.4 \, \text{mol/kg}$), we can solve for $K_b$. Rearranging the formula gives $K_b = \frac{\Delta T_b}{i \cdot m} = \frac{1.23 °C}{1 \cdot 0.4 \, \text{mol/kg}} = 3.08 °C/\text{molal}$.

Question 18

Benzene boils at 80.10 ° C and has a molal boiling constant, k_b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molality of the solute?

Accepted Answer

First, calculate the change in boiling point (ΔTb) using the formula:

ΔTb = Tb(solution) - Tb(solvent)
ΔTb = 81.10 °C - 80.10 °C = 1.00 °C

Next, calculate the molality (m) of the solute using the formula:

ΔTb = kbm
m = ΔTb / kb

where kb is the molal boiling constant for benzene (2.53 °C/m).

m = 1.00 °C / (2.53 °C/m)
m = 0.395 m

Therefore, the molality of the solute is 0.395 m, which corresponds to choice C.

Question 19

What is the vapor pressure lowering D P of an aqueous glucose solution prepared by dissolving 9.01 grams of glucose, C₆H₁₂O₆, in 17.1 grams of water? Molar Mass(C₆H₁₂O₆) = 180.16 g/mol and Molar Mass(H₂O) = 18.02 g/mol; vapor pressure of pure water = 23.8 torr (at 25 ° C)

Accepted Answer

To calculate the vapor pressure lowering ($ \Delta P $), we use Raoult's law, which states that the vapor pressure lowering is directly proportional to the mole fraction of the solute in the solution. First, we calculate the moles of glucose and water:- Moles of glucose = mass / molar mass = $9.01 \, \text{g} / 180.16 \, \text{g/mol} = 0.05 \, \text{mol}$- Moles of water = mass / molar mass = $17.1 \, \text{g} / 18.02 \, \text{g/mol} = 0.95 \, \text{mol}$The mole fraction of glucose ($X_{\text{glucose}}$) is calculated as the moles of glucose divided by the total moles of the solution:$$X_{\text{glucose}} = \frac{0.05}{0.05 + 0.95} = \frac{0.05}{1} = 0.05$$The vapor pressure lowering ($ \Delta P $) is then the product of the mole fraction of the solute and the vapor pressure of the pure solvent:$$\Delta P = X_{\text{glucose}} \times \text{vapor pressure of pure water} = 0.05 \times 23.8 \, \text{torr} = 1.19 \, \text{torr}$$Rounding to two decimal places, $ \Delta P = 1.20 \, \text{torr}$.

Question 20

A solution that contains 3.00 g of urea [CO(NH₂)₂] dissolved in 78.1 g of benzene (C₆H₆) has a freezing point of 2.37 ° C. The normal freezing point of benzene is 5.51 ° C. What is the freezing point depression constant for this solvent?

Accepted Answer

The freezing point depression (ΔTf) is calculated using the formula ΔTf = i * Kf * m, where i is the van't Hoff factor (1 for urea), Kf is the freezing point depression constant, and m is the molality. The change in freezing point is 5.51 °C - 2.37 °C = 3.14 °C. The molality (m) is calculated as moles of solute per kg of solvent. Moles of urea = 3.00 g / 60.06 g/mol = 0.0499 mol. Molality = 0.0499 mol / 0.0781 kg = 0.639 m. Solving for Kf gives Kf = ΔTf / m = 3.14 °C / 0.639 m = 4.91 °C/m, which is closest to 4.90 °C/m.

Quiz 12: Solutions

A sample of 2.00 g of the non-volatile compound urea, CO(NH₂) ₂ (molar mass = 60 g/mol) , is dissolved in 20.0 grams of water. What is the vapor pressure of water above this solution? The vapor pressure of pure water at the temperature of the experiment is 0.150 atm.

A 10.0 mass percent solution of a nonelectrolyte dissolved in carbon tetrachloride has a boiling point that is 4.50 ° C higher than the solvent. What is the molality of the solute given k_b is 5.03 ° C/m?

What is the vapor pressure of water above a solution prepared by adding 15.0 grams of lactose (C₁₂H₂₂O₁₁) to 100.0 grams of water at 338 K? The vapor pressure of pure water at 338 K equals 187.5 torr. Molar Mass(C₁₂H₂₂O₁₁) = 342.30 g/mol and Molar Mass(H₂O) = 18.015 g/mol.

A 10.0 mass percent solution of a nonelectrolyte dissolved in carbon tetrachloride has a boiling point that is 4.50 ° C higher than the solvent. What is the molar mass of this solute given k_b is 5.03 ° C/m?

The boiling point of pure benzene (C₆H₆) is 80.10 ° C. A solution of 7.39 g of a solid dissolved in 85.0 g of benzene boils at 82.49 ° C. If k_b for benzene is 2.53 ° C/molal, the molar mass of the solid is:

What is the vapor pressure of water above an aqueous solution prepared by mixing 300 grams of glycerol (non-volatile solute) and 505 grams of H₂O at 25 ° C? Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H₂O) = 18.02 g/mol; P_H2O(25 ° C) = 23.76 torr

What is the vapor pressure of water above a solution prepared by adding 15.0 grams of lactose (C₁₂H₂₂O₁₁) to 100.0 grams of water at 338 K. Molar Mass(C₁₂H₂₂O₁₁) = 342.3 g/mol and Molar Mass(H₂O) = 18.015 g/mol; P ° _H2O(338 K) = 187.5 torr.

Which expression shown below is correct for solving for the freezing temperature of a solution?

Benzene boils at 80.10 ° C and has a molal boiling constant, k_b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molar mass of the solute?

What is the vapor pressure of water above a solution prepared by adding 0.750 moles of lactose tO₅ .55 moles of water at 338 K? The vapor pressure of pure water at 338 K equals 187.5 torr.

Ethylene glycol is mixed with water and added to an automobile's radiator. Which of the following occurs? I. The freezing temperature of the mixture rises. II. The vapor pressure inside the radiator increases. III. The boiling temperature of the mixture rises.

What is the vapor pressure of a solution of 30.0 g glycerol (C₃H₈O₃) in 504.5 grams of water at 25 ° C. The vapor pressure of pure water at 25 ° C is 23.76 torr and Molar Mass(C₃H₈O₃) = 92.09 g/mol.

A sample of 20.0 g of a nonvolatile, nonelectrolyte (molar mass = 120 g/mol) is dissolved in 10.0 grams of ethanol (C₂H₅OH) . If the vapor pressure of pure ethanol at the temperature of the experiment is 0.250 atm, what is the vapor pressure of ethanol above the solution?

At 120 ° C, the vapor pressure of pure chlorobenzene (C₆H₅Cl) is 559 torr. What is the vapor pressure of a solution of 5.00 g of naphthalene (C₁₀H₈) in 50.0 g of chlorobenzene? (Assume that naphthalene is not volatile.)

A solution that contains 2.00 g of ferrocene [molar mass = 186.0 g/mol] dissolved in 50.0 g of ethylene bromide (C₂H₄Br₂) has a freezing point of 7.26 ° C. The normal freezing point of ethylene bromide is 9.80 ° C. What is the freezing point depression constant for this solvent?

What is the vapor pressure lowering , D P, for an aqueous solution prepared by mixing 50.0 grams of glycerol (non-volatile solute) and 500 grams of H₂O at 25 ° C? Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H₂O) = 18.02 g/mol; P_H2O(25 ° C) = 23.76 torr.

The boiling point of pure acetic acid (C₂H₄O₂) is 117.90 ° C. A solution of 1.80 g of glucose (molar mass = 180 g/mol) dissolved in 25.0 g acetic acid was found to boil at 119.13 ° C. The value of k_bfor acetic acid (in units of ° C/molal) is:

Benzene boils at 80.10 ° C and has a molal boiling constant, k_b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molality of the solute?

What is the vapor pressure lowering D P of an aqueous glucose solution prepared by dissolving 9.01 grams of glucose, C₆H₁₂O₆, in 17.1 grams of water? Molar Mass(C₆H₁₂O₆) = 180.16 g/mol and Molar Mass(H₂O) = 18.02 g/mol; vapor pressure of pure water = 23.8 torr (at 25 ° C)

A solution that contains 3.00 g of urea [CO(NH₂) ₂] dissolved in 78.1 g of benzene (C₆H₆) has a freezing point of 2.37 ° C. The normal freezing point of benzene is 5.51 ° C. What is the freezing point depression constant for this solvent?

Quiz 12: Solutions

A sample of 2.00 g of the non-volatile compound urea, CO(NH2) 2 (molar mass = 60 g/mol) , is dissolved in 20.0 grams of water. What is the vapor pressure of water above this solution? The vapor pressure of pure water at the temperature of the experiment is 0.150 atm.

A 10.0 mass percent solution of a nonelectrolyte dissolved in carbon tetrachloride has a boiling point that is 4.50 ° C higher than the solvent. What is the molality of the solute given kb is 5.03 ° C/m?

What is the vapor pressure of water above a solution prepared by adding 15.0 grams of lactose (C12H22O11) to 100.0 grams of water at 338 K? The vapor pressure of pure water at 338 K equals 187.5 torr. Molar Mass(C12H22O11) = 342.30 g/mol and Molar Mass(H2O) = 18.015 g/mol.

A 10.0 mass percent solution of a nonelectrolyte dissolved in carbon tetrachloride has a boiling point that is 4.50 ° C higher than the solvent. What is the molar mass of this solute given kb is 5.03 ° C/m?

The boiling point of pure benzene (C6H6) is 80.10 ° C. A solution of 7.39 g of a solid dissolved in 85.0 g of benzene boils at 82.49 ° C. If kb for benzene is 2.53 ° C/molal, the molar mass of the solid is:

What is the vapor pressure of water above an aqueous solution prepared by mixing 300 grams of glycerol (non-volatile solute) and 505 grams of H2O at 25 ° C? Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H2O) = 18.02 g/mol; PH2O (25 ° C) = 23.76 torr

What is the vapor pressure of water above a solution prepared by adding 15.0 grams of lactose (C12H22O11) to 100.0 grams of water at 338 K. Molar Mass(C12H22O11) = 342.3 g/mol and Molar Mass(H2O) = 18.015 g/mol; P ° H2O(338 K) = 187.5 torr.

Which expression shown below is correct for solving for the freezing temperature of a solution?

Benzene boils at 80.10 ° C and has a molal boiling constant, kb, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molar mass of the solute?

What is the vapor pressure of water above a solution prepared by adding 0.750 moles of lactose tO5 .55 moles of water at 338 K? The vapor pressure of pure water at 338 K equals 187.5 torr.

Ethylene glycol is mixed with water and added to an automobile's radiator. Which of the following occurs? I. The freezing temperature of the mixture rises. II. The vapor pressure inside the radiator increases. III. The boiling temperature of the mixture rises.

What is the vapor pressure of a solution of 30.0 g glycerol (C3H8O3) in 504.5 grams of water at 25 ° C. The vapor pressure of pure water at 25 ° C is 23.76 torr and Molar Mass(C3H8O3) = 92.09 g/mol.

A sample of 20.0 g of a nonvolatile, nonelectrolyte (molar mass = 120 g/mol) is dissolved in 10.0 grams of ethanol (C2H5OH) . If the vapor pressure of pure ethanol at the temperature of the experiment is 0.250 atm, what is the vapor pressure of ethanol above the solution?

At 120 ° C, the vapor pressure of pure chlorobenzene (C6H5Cl) is 559 torr. What is the vapor pressure of a solution of 5.00 g of naphthalene (C10H8) in 50.0 g of chlorobenzene? (Assume that naphthalene is not volatile.)

A solution that contains 2.00 g of ferrocene [molar mass = 186.0 g/mol] dissolved in 50.0 g of ethylene bromide (C2H4Br2) has a freezing point of 7.26 ° C. The normal freezing point of ethylene bromide is 9.80 ° C. What is the freezing point depression constant for this solvent?

What is the vapor pressure lowering , D P, for an aqueous solution prepared by mixing 50.0 grams of glycerol (non-volatile solute) and 500 grams of H2O at 25 ° C? Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H2O) = 18.02 g/mol; PH2O (25 ° C) = 23.76 torr.

The boiling point of pure acetic acid (C2H4O2) is 117.90 ° C. A solution of 1.80 g of glucose (molar mass = 180 g/mol) dissolved in 25.0 g acetic acid was found to boil at 119.13 ° C. The value of kb for acetic acid (in units of ° C/molal) is:

Benzene boils at 80.10 ° C and has a molal boiling constant, kb, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molality of the solute?

What is the vapor pressure lowering D P of an aqueous glucose solution prepared by dissolving 9.01 grams of glucose, C6H12O6, in 17.1 grams of water? Molar Mass(C6H12O6) = 180.16 g/mol and Molar Mass(H2O) = 18.02 g/mol; vapor pressure of pure water = 23.8 torr (at 25 ° C)

A solution that contains 3.00 g of urea [CO(NH2) 2] dissolved in 78.1 g of benzene (C6H6) has a freezing point of 2.37 ° C. The normal freezing point of benzene is 5.51 ° C. What is the freezing point depression constant for this solvent?

A sample of 2.00 g of the non-volatile compound urea, CO(NH₂) ₂ (molar mass = 60 g/mol) , is dissolved in 20.0 grams of water. What is the vapor pressure of water above this solution? The vapor pressure of pure water at the temperature of the experiment is 0.150 atm.

A 10.0 mass percent solution of a nonelectrolyte dissolved in carbon tetrachloride has a boiling point that is 4.50 ° C higher than the solvent. What is the molality of the solute given k_b is 5.03 ° C/m?

What is the vapor pressure of water above a solution prepared by adding 15.0 grams of lactose (C₁₂H₂₂O₁₁) to 100.0 grams of water at 338 K? The vapor pressure of pure water at 338 K equals 187.5 torr. Molar Mass(C₁₂H₂₂O₁₁) = 342.30 g/mol and Molar Mass(H₂O) = 18.015 g/mol.

A 10.0 mass percent solution of a nonelectrolyte dissolved in carbon tetrachloride has a boiling point that is 4.50 ° C higher than the solvent. What is the molar mass of this solute given k_b is 5.03 ° C/m?

The boiling point of pure benzene (C₆H₆) is 80.10 ° C. A solution of 7.39 g of a solid dissolved in 85.0 g of benzene boils at 82.49 ° C. If k_b for benzene is 2.53 ° C/molal, the molar mass of the solid is:

What is the vapor pressure of water above an aqueous solution prepared by mixing 300 grams of glycerol (non-volatile solute) and 505 grams of H₂O at 25 ° C? Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H₂O) = 18.02 g/mol; P_H2O(25 ° C) = 23.76 torr

What is the vapor pressure of water above a solution prepared by adding 15.0 grams of lactose (C₁₂H₂₂O₁₁) to 100.0 grams of water at 338 K. Molar Mass(C₁₂H₂₂O₁₁) = 342.3 g/mol and Molar Mass(H₂O) = 18.015 g/mol; P ° _H2O(338 K) = 187.5 torr.

Benzene boils at 80.10 ° C and has a molal boiling constant, k_b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molar mass of the solute?

What is the vapor pressure of water above a solution prepared by adding 0.750 moles of lactose tO₅ .55 moles of water at 338 K? The vapor pressure of pure water at 338 K equals 187.5 torr.

What is the vapor pressure of a solution of 30.0 g glycerol (C₃H₈O₃) in 504.5 grams of water at 25 ° C. The vapor pressure of pure water at 25 ° C is 23.76 torr and Molar Mass(C₃H₈O₃) = 92.09 g/mol.

A sample of 20.0 g of a nonvolatile, nonelectrolyte (molar mass = 120 g/mol) is dissolved in 10.0 grams of ethanol (C₂H₅OH) . If the vapor pressure of pure ethanol at the temperature of the experiment is 0.250 atm, what is the vapor pressure of ethanol above the solution?

At 120 ° C, the vapor pressure of pure chlorobenzene (C₆H₅Cl) is 559 torr. What is the vapor pressure of a solution of 5.00 g of naphthalene (C₁₀H₈) in 50.0 g of chlorobenzene? (Assume that naphthalene is not volatile.)

A solution that contains 2.00 g of ferrocene [molar mass = 186.0 g/mol] dissolved in 50.0 g of ethylene bromide (C₂H₄Br₂) has a freezing point of 7.26 ° C. The normal freezing point of ethylene bromide is 9.80 ° C. What is the freezing point depression constant for this solvent?

What is the vapor pressure lowering , D P, for an aqueous solution prepared by mixing 50.0 grams of glycerol (non-volatile solute) and 500 grams of H₂O at 25 ° C? Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H₂O) = 18.02 g/mol; P_H2O(25 ° C) = 23.76 torr.

The boiling point of pure acetic acid (C₂H₄O₂) is 117.90 ° C. A solution of 1.80 g of glucose (molar mass = 180 g/mol) dissolved in 25.0 g acetic acid was found to boil at 119.13 ° C. The value of k_bfor acetic acid (in units of ° C/molal) is:

Benzene boils at 80.10 ° C and has a molal boiling constant, k_b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molality of the solute?

What is the vapor pressure lowering D P of an aqueous glucose solution prepared by dissolving 9.01 grams of glucose, C₆H₁₂O₆, in 17.1 grams of water? Molar Mass(C₆H₁₂O₆) = 180.16 g/mol and Molar Mass(H₂O) = 18.02 g/mol; vapor pressure of pure water = 23.8 torr (at 25 ° C)

A solution that contains 3.00 g of urea [CO(NH₂) ₂] dissolved in 78.1 g of benzene (C₆H₆) has a freezing point of 2.37 ° C. The normal freezing point of benzene is 5.51 ° C. What is the freezing point depression constant for this solvent?