Question

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.

Answer 1

The 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.

Answer 2

that to*

Answer 3

And be sure that you make the molefraction equal to the ratio of moles and not the ratio of masses of the two substances.

Answer 4

Enough hints or need a bit more?

Answer 5

I found mole fraction be 0.563. However, for some reason it is wrong.

Answer 6

What I did was that I let "x" equal the grams of each substance. So x g of CCl4 = x g of CHCl3

Answer 7

Then 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

Answer 8

x/119.368 mol, x/153.81 mol

Answer 9

Then I realized that the 'x' cancel out when put into the mole fraction.

Answer 10

You can factor them out.

Answer 11

Then 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

Answer 12

I think you are referencing the wrong problem.

Answer 13

Then I did this:
P_total - P_CCl4
------------
(PCHCl3)

= X_CHCl3

Answer 14

OOPS!

Answer 15

Ok I need another hint.

Answer 16

Try 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?

Answer 17

Xishem, 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

Answer 18

From here, I do not know how to proceed

Answer 19

\[mass_{sol'n}=mass_{urea}+mass_{H_2O}\]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.

Answer 20

I got 0.185mol Urea and 2.26mol water.

Answer 21

That seems right, I wonder what mistake I made.

Answer 22

My moles of water was wrong.

Answer 23

So you got the moles of urea correct?

Answer 24

Oh, never mind, I realize.

Answer 25

I 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)

Answer 26

Sorry not moles urea, rather grams of urea I am getitng that much of.

Answer 27

Anyways, 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.

Answer 28

The rest is just a math exercise.

Answer 29

My 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.

Answer 30

Rogue's solution is much more elegant and easier to execute.