Partly ok but not completely.
First I see where you made a typo with '
p = 0.041(17.54) and you wrote 0.41. (But that's just a typo because the 0.726 number is ok.
Second, you didn't finish the problem. You have calculated the partial pressure of the water but the problem asks for the "lowering." So you must subtract this number from 17.54 to find the lowering.
Third problem is for AlCl3. You apparently used 3 for i although it isn't labeled as such. Wouldn't i be 4 for AlCl3? Also for this you need to subtract from 17.54 to find lowering.
There is a shorcut you may use. When calculating "lowering" you can use
delta Psolvent = Xsolute*Posolvent
Given that the vapor pressure of water is 17.54 torr at 20 C, calculate the vapor-pressure lowering of aqueous solutions that are 2.40 m in (a) sucrose, C12H22O11, and (b) aluminum chloride. Assume 100% dissociation for electrolytes.
Sucrose Delta P = __ torr
Aluminum Chloride Delta P = __ torr
here's what I did..
2.4/(2.4+55.56) = 0.041
p = (0.41)(17.54) = 0.726 torr sucrose?
p = (3)(0.726) = 2.178 torr for aluminum chloride?
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