How would you determine the vapor pressure of a solution at 25C that contains 76.6 g of glucose (C6H12O6) in 250.0 mL of water? The vapor pressure of pure water at 25C is 23.8 torr.

Answer 1

#"23.1 torr"#

The first thing to do here is determine the mass of the sample of water by using water's density at #25^@"C"#, which is equal to
#rho = "0.99705 g/mL"#

Water Density (https://tutor.hix.ai)

This implies that the water's mass will be

#250.0color(red)(cancel(color(black)("mL"))) * "0.99705 g"/(1color(red)(cancel(color(black)("mL")))) = "249.26 g"#

Since glucose is a non-volatile substance, the only factors affecting the solution's vapor pressure are the mole fraction of water and the vapor pressure of pure water.

To put it simply, only water vapor will be present at the solution's vapor pressure.

Now, calculate how many moles of each you have using the molar masses of glucose and water.

#76.6color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156color(red)(cancel(color(black)("g")))) = "0.4252 moles glucose"#

additionally

#249.26color(red)(cancel(color(black)("g"))) * "1 mole water"/(18.015color(red)(cancel(color(black)("g")))) = "13.836 moles water"#

The ratio of the amount of water to the total amount of moles in the solution is known as the mole fraction of water.

There will be a total of moles in the solution.

#n_"total" = n_"glucose" + n_"water"#
#n_"total" = 0.4252 + 13.836 = "14.261 moles"#

This implies that the water mole fraction will be

#chi_"water" = (13.836color(red)(cancel(color(black)("moles"))))/(14.261color(red)(cancel(color(black)("moles")))) = 0.9702#

Consequently, the solution's vapor pressure will be

#P_"sol" = chi_"water" * P_"water"^@#
#P_"sol" = 0.9702 * "23.8 torr" = "23.091 torr"#

The response, rounded to three sig figs, is

#P_"sol" = color(green)("23.1 torr")#
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Answer 2

To determine the vapor pressure of the solution, you would use Raoult's Law, which states that the vapor pressure of a solution is proportional to the mole fraction of solvent present. First, calculate the mole fraction of water in the solution. Then, use the formula:

Vapor pressure of solution = Mole fraction of solvent * Vapor pressure of pure solvent

To find the mole fraction of water:

  1. Calculate the moles of glucose using its molar mass.
  2. Calculate the moles of water using its volume and molar mass.
  3. Add the moles of glucose and water together.
  4. Divide the moles of water by the total moles to find the mole fraction of water.

Then, substitute the mole fraction of water and the vapor pressure of pure water into the formula to find the vapor pressure of the solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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