Colligative Property Question (Almost finished need last step help)?

How many g of AlCl3 would have to be dissolved in 150 mL of water to give a solution that has a vapor pressure of 37.4 torr at 35°C? Assume complete dissociation of the solute and ideal solution behavior. At 35°C, the vapor pressure of pure water is 42.2 torr.

I started the equation by filling in the

Psolution= X solvent * P

and using "x" as a variable for the number of moles of solute

37.4= 8.33/8.33+ x * 42.2

Which I carried out equations to simply to:

8.33= 0.886x + 7.382

And further to

x = 1.07 moles

Now that I have moles I'm not sure how to set up to solve for grams of AlCl3 because in the problem it states that it completely dissociates and I believe there is something about colligative property questions having their answer solved to the particles but I'm unsure of how to set that up and end up getting grams.

Answer:
You are close

Since AlCl3 dissociates into 4 moles of particles, use 4x instead of just x

Xsolvent = moles H2O/(4x moles AlCl3 + moles H2O)

X solvent = 8.32/(4 x moles AlCl3 + 8.32)

Then solve for moles of AlCl3 and convert to grams

The answers post by the user, for information only, FunQA.com does not guarantee the right.

FunQA.com

Colligative Property Question (Almost finished need last step help)?

Topic