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The vapor pressure of diethyl ether is 463.57 mm Hg at 25 degrees celsius. How many grams of aspirin C9H8O4, a non volatile non...Asked by Hannah
The vapor pressure of diethyl ether is 463.57 mm Hg at 25 degrees celsius. How many grams of aspirin C9H8O4, a non volatile nonelectrolyte (mw=180.1g/mol), must be added to 216.7g of ether to reduce the vapor pressure to 453.74 mm Hg.
Ether=74.12g/mol
so I did 463.57 - 453.74 = 7.600
Then I found moles of ether to be 4.0313.
I do not understand the next step to find Xether.
Ether=74.12g/mol
so I did 463.57 - 453.74 = 7.600
Then I found moles of ether to be 4.0313.
I do not understand the next step to find Xether.
Answers
Answered by
DrBob222
You had better subtract again. 7.600 is not right.
You need the mole fraction of ether. You get that by
pether = Xether*Po ether
You know partial pressure of ether to be 453.74 (that's what you want it to be)
You know Po ether to be 463.57. Solve for Xether.
You need the mole fraction of ether. You get that by
pether = Xether*Po ether
You know partial pressure of ether to be 453.74 (that's what you want it to be)
You know Po ether to be 463.57. Solve for Xether.
Answered by
Hannah
I subtracted again and got 9.83.
So do I set it up like this
453.74= Xether * 463.57
So do I set it up like this
453.74= Xether * 463.57
Answered by
Hannah
Disregard question.
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