[student8607]

How many grams of NaBr must be added to 250g of water to lower the vapor pressure by 1.30mmHg assuming complete dissociation? The vapor pressure of water is 55.3mmHg.

OK, so the P_soln = P_solv times X_solv

1.30mmHg for P of solution = [55.3mmHg for P of solvent] times [?mole fraction]

^Can I use 1.30 as P for solution? It says "lowers by" but what is it originally?
I believe I am supposed to solve for mole fraction to get grams of solvent

[student8607]

 

[student8607]

0.01175 = n_solute / (250/18)
n_solute = 0.1632 moles

The weight of NaBr (solute) to be added = 0.1632 * formula. wt of NaBr = 0.1632 * 103 = 16.81 g

Very informative and helpful post.

Just confused in the mole fraction. Where does the 250/18 come from? That is to be total moles in the solution, right?