A glucose solution contains 55.8 g of glucose (C 6 H 12 O 6 ) in 455 g of water. Determine the freezing point and boiling point of the solution.

Hi everyone here we have a question telling us that a mass of 66.5 g of Ben's OIC acid is involved in 545 g of water, calculate the boiling and freezing point of the resulting solution. So we first need to calculate the morality. The morality equals the moles of the solute over the kilograms of the solvent. The molar mass Equals 12.01 grams per mole times seven Plus 1.008 g per mole times six plus 16 zero g per mole times two, Which equals 122. g per mole. The moles of Ben's OIC acid Equals 66.5 g times one mole, Divided by the Molar Mass, which is 122.118g, Which equals 0. moles of Ben's OIC acid. The mass of water equals g of water times one kg Over 10 to the 3rd g. And our g are going to cancel out leaving us with 0.545 kg. So our morality Equals 0.544 h five moles. A Benz OIC acid Divided by 0. kilograms of water, which equals 0.999 to morality, Benzoate acid is a non electrolyte, Which means it's fan Hoff factor is one. So now, to calculate the boiling point of our solution, our change in boiling point equals our boiling point of our solution minus the boiling point of our peer solvent. Our change and boiling point equals R. Van Hoff factor times our bowling point constant times our morality, our boiling point constant Is 0.512 degrees Celsius per morality. Our boiling point of our solvent Is 100°C. So our change in our boiling point equals one Time 0.512°C per morality Times 0. to morality Equals 0.5116°C. Our boiling point of our solution equals the change in boiling point plus the boiling point of our solvent, Which equals 100°C plus 0.5116 degrees Celsius, which equals .512°C. And that is the answer for our first part. Next we're going to calculate the freezing point, so the freezing point equals the freezing point of our pure solvent minus the freezing point of our solution. Our change in freezing point equals our van Hoff factor times our freezing point constant times our morality Are freezing point constant equals 1.86°C per morality. The freezing point for our solvent Is 0°C. So the change in our freezing point equals one Times 1.86 degrees Celsius per morality Times 0. Morality Equals 1.858 degrees Celsius. Our freezing point of our solution equals the freezing point of our solvent minus our change in freezing point, Which equals 0°C -1.858°C, which equals negative 1.858°C. And that is our second and final answer. Thank you for watching. Bye.

Ch.13 - Solutions

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Tro 4th Edition

Ch.13 - Solutions

Problem 77

Chapter 13, Problem 77

A glucose solution contains 55.8 g of glucose (C₆H₁₂O₆) in 455 g of water. Determine the freezing point and boiling point of the solution.

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Identify the colligative properties involved: freezing point depression and boiling point elevation.

Calculate the molality (m) of the solution using the formula: \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). First, find the moles of glucose by dividing the mass of glucose by its molar mass.

Use the freezing point depression formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor (1 for glucose), \( K_f \) is the cryoscopic constant for water, and \( m \) is the molality.

Use the boiling point elevation formula: \( \Delta T_b = i \cdot K_b \cdot m \), where \( K_b \) is the ebullioscopic constant for water.

Determine the new freezing and boiling points by subtracting \( \Delta T_f \) from the normal freezing point of water (0°C) and adding \( \Delta T_b \) to the normal boiling point of water (100°C).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Colligative Properties

Colligative properties are physical properties of solutions that depend on the number of solute particles in a given amount of solvent, rather than the identity of the solute. These properties include boiling point elevation and freezing point depression, which occur when a solute is added to a solvent, affecting the solvent's phase changes.

Freezing Point Depression

Freezing point depression is a colligative property that describes the decrease in the freezing point of a solvent when a solute is dissolved in it. The extent of freezing point depression can be calculated using the formula ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van 't Hoff factor, Kf is the freezing point depression constant, and m is the molality of the solution.

Boiling Point Elevation

Boiling point elevation is another colligative property that refers to the increase in the boiling point of a solvent when a solute is added. This phenomenon can be quantified using the formula ΔTb = i * Kb * m, where ΔTb is the change in boiling point, i is the van 't Hoff factor, Kb is the boiling point elevation constant, and m is the molality of the solution.

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Textbook Question

A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 258 torr. Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature. What is the mole fraction composition of the mixture? (Assume ideal behavior.)

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