[ethan]

The vapor pressure of liquid A (mol wt 120) is 70 mmHg at 25 Celsius. What is the vapor pressure of a solution made by dissolving 10g C6H4Cl2 in 30g of A at 25 Celsius?

Moles of A = 0.25 mol
Moles of C6H4Cl2 = 0.068 mol

Raoult's Law...

70 mmHg = (0.25 mol solvent)/(0.318 mol) x Psolv

Answer: 89 mmHg?

[Borek]

Higher pressure after dissolving something?

[ethan]

Am I not supposed to use Raoults Law?

[Borek]

Raoult's Law is OK, but you did something wrong and it is obvious without checking your calculations. Raoult's Law states, that vapor pressure over solution is proportional to molar fraction of the substance. When you add something non volatile, molar fraction of solvent (A in your case) becomes less then 1 - so the A pressure must be lower than initial value. 89 mm Hg is larger than 70 mm Hg, so it is obvious you did something wrong.

[ethan]

Oh, okay.

P1 = (.25mol/.318mol) x 70mmHg = 55.03 mmHg

But then dont I also need the Vapor Pressure of the Dichlorobenzene at 25 degrees Celsius? But that isn't given in the problem...

[Borek]

Seems OK. You are either to assume dichlorobenze to be non volatile - or to check its vapor pressure in CRC handbook or something similar. On the HS level most likely first approach will be OK, especially if you were not told if it is ortho, meta or para, as their properties differ.

[ethan]

Oh okay thank you :]