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anonymous
 4 years ago
Find the mass of urea CH4N2O needed to prepare 51.8g of a solution in water in which the mole fraction of urea is 7.55×10−2.
I would like a hint. Not the solution please.
anonymous
 4 years ago
Find the mass of urea CH4N2O needed to prepare 51.8g of a solution in water in which the mole fraction of urea is 7.55×10−2. I would like a hint. Not the solution please.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The mole fraction tells you the percentage of urea in the solution. So you can use that you calculate the mass that would make that percent, out of 51.8 g.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And be sure that you make the molefraction equal to the ratio of moles and not the ratio of masses of the two substances.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Enough hints or need a bit more?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I found mole fraction be 0.563. However, for some reason it is wrong.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What I did was that I let "x" equal the grams of each substance. So x g of CCl4 = x g of CHCl3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then I multiplied this x value of grams for each individually by their molar masses of each individual substance to get the moles of each substance

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x/119.368 mol, x/153.81 mol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then I realized that the 'x' cancel out when put into the mole fraction.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You can factor them out.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then what I did was I found the mole fractions and for each substance individually, I found its vapour pressure above the solution. Then I did P_total = P_CCl4 + P_CHCl3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think you are referencing the wrong problem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then I did this: P_total  P_CCl4  (PCHCl3) = X_CHCl3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok I need another hint.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try assuming 1mol of solution. How many moles of urea and how many mols of water would have to be present to give the correct molefraction?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Xishem, I forgot to include that in my opening post. I actually did come to that and this is what I got: per 1 mole of solution, you have 0.0755 mol CH4N2O, 0.9245 mol H2O

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0From here, I do not know how to proceed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[mass_{sol'n}=mass_{urea}+mass_{H_2O}\]Find the ratios of masses from the assumed 1mol of sol'n, and then express one of the masses in terms of the other mass. I think this should work.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got 0.185mol Urea and 2.26mol water.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That seems right, I wonder what mistake I made.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My moles of water was wrong.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So you got the moles of urea correct?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, never mind, I realize.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got 0.185mol Urea and 2.26mol water. ^ I am getting 4.53mol urea by doing this: mass urea = (0.0755mol)*(60.07g/1mol urea)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry not moles urea, rather grams of urea I am getitng that much of.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Anyways, here is how you solve it, warrior. The molar mass of urea is 60.0553 g/mol. The molar mass of water is 18.0153 g/mol. Let u = moles of urea. Let w = moles of water.\[\frac {u}{u+w} = 0.0755\]\[51.8g = 60.0553u + 18.0153w\] Solve for the system of equations.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The rest is just a math exercise.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My solution was this. Assume 1 mol of sol'n, then:\[1 mol(sol'n)=0.9245mol(H_2O)+0.0755mol(urea)\]Then I found the ratio of mols of water to mols of urea:\[\frac{0.9245}{0.0755}=12.245\] Next, I used this expression:\[mass_{sol'n}=mass_{H_2O}+mass_{urea}\]And since...\[mass_{compound}=MolarMass_{compound}*mols_{compound}\]You can do some substitution to change the above equation (mass sol'n...) into:\[mass_{sol'n}=(mols_{H_2O}*MM_{H_2O})+(mols_{urea}*MM_{urea})\] We know that for this solution there are 12.245 times as many mols of water as there are of urea, so...\[1mol(urea)=12.245mol(H_2O)\] Now, let's express this ratio in terms of x within the mass of solution equation: \[mass_{sol'n}=(12.245)(x)*(MM_{H_2O})+(1)(x)*(MM_{urea})\]Now, let's plug in some numbers and see what we get...\[51.8g=(12.245)(x)*(18.016g*mol^{1})+(x)*(60.062g*mol^{1})\]After some algebra, we can find that...\[x=0.1846\] If we look at the original equation, we can see that x is the same thing as the number of mols of urea, so we know that there are 0.1845mol urea. From here, you can convert this to grams. After you convert it to grams, you can check your answer by checking to see that the molefraction is 7.55 x 10^2, which indeed it is.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Rogue's solution is much more elegant and easier to execute.
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