Asked by Kelley
Find the mass of urea needed to prepare 50.1 of a solution in water in which the mole fraction of urea is 7.50×10−2.
Answers
Answered by
DrBob222
50.1 what?
Answered by
Kelley
grams.
Answered by
DrBob222
Let x = mass urea.
n urea = x/60.06
n H2O = 50.1-(x/18)
n urea = [n urea/(n urea + n H2O)]
So (x/60)/[x/60)+{50.1-x}/18] = 0.075 and solve for x.
If I didn't make a math error the answer is approximately 10 g.
n urea = x/60.06
n H2O = 50.1-(x/18)
n urea = [n urea/(n urea + n H2O)]
So (x/60)/[x/60)+{50.1-x}/18] = 0.075 and solve for x.
If I didn't make a math error the answer is approximately 10 g.
Answered by
Anonymous
where did you get 60.06 from?
Answered by
Anonymous
oh, molar mass, duh. got it
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