How do you calculate the vapor pressure of water above a solution prepared by dissolving 39.5 g in glycerin (#C_3H_8O_3#) to 149 g of water at 343 K?

Answer 1

The vapour pressure of water at #343*K# is #233.8*mm*Hg.....#The vapour pressure of your solution is #222.3*mm*Hg.#

The given vapour pressure was taken from this site here. (I am doing you a favour here by digging for data that you should have supplied yourself!)

Now it is a fact that the vapour pressure of a solution is proportional to the mole fraction of the volatile component(s) of the solution. For this problem, we assume that glycerine is involatile (this is not entirely true but for the purposes of this problem I am not going to pursue more info).

So we work out the #"mole fraction"# of each component:
#chi_"water"="Moles of water"/"Moles of water + moles of glycerol"#
#chi_"water"=[[149*g)/(18.01*g*mol^-1)]/([149*g)/(18.01*g*mol^-1)+(39.5*g)/(92.09*g*mol^-1]#
#chi_"water"=0.951#. Because there are only the two components in solution, #chi_"glycerol"=1-chi_"water"=1-0.951=0.0490#.

And so (finally), we can calculate the vapour pressure of the solution, which is proportional to its mole fraction of the volatile components, i.e.

#"Vapour pressure of solution "=#
#chi_"water"xx"vapour pressure of pure water"=0.951xx233.8*mm*Hg=??"mm Hg".#

This is, as we would expect, slightly depressed from the vapour pressure of pure water.

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Answer 2

To calculate the vapor pressure of water above the solution, you can use Raoult's Law, which states that the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent in the solution. First, calculate the mole fraction of water:

Moles of water = mass of water (in grams) / molar mass of water Moles of glycerin = mass of glycerin (in grams) / molar mass of glycerin

Then, calculate the mole fraction of water: Mole fraction of water = Moles of water / (Moles of water + Moles of glycerin)

Finally, use Raoult's Law to calculate the vapor pressure of water: Vapor pressure of water = mole fraction of water * vapor pressure of pure water at the given temperature.

Ensure the vapor pressure of water is given or can be looked up at 343 K. Once you have that value, plug in the calculated mole fraction of water to find the vapor pressure above 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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