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.

Question

Accepted Answer

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.

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.

Verified Solution

Key Concepts

Colligative Properties

Freezing Point Depression

Boiling Point Elevation

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.

Verified Solution

Key Concepts

Colligative Properties

Freezing Point Depression

Boiling Point Elevation

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.