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

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