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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.

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1. Rogue

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.

2. Rogue

that to*

3. Xishem

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

4. Rogue

Enough hints or need a bit more?

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

x/119.368 mol, x/153.81 mol

9. anonymous

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

10. anonymous

You can factor them out.

11. anonymous

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

12. Xishem

I think you are referencing the wrong problem.

13. anonymous

Then I did this: P_total - P_CCl4 ------------ (PCHCl3) = X_CHCl3

14. anonymous

OOPS!

15. anonymous

Ok I need another hint.

16. Xishem

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?

17. anonymous

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

18. anonymous

From here, I do not know how to proceed

19. Xishem

$mass_{sol'n}=mass_{urea}+mass_{H_2O}$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.

20. Xishem

I got 0.185mol Urea and 2.26mol water.

21. Rogue

That seems right, I wonder what mistake I made.

22. Rogue

My moles of water was wrong.

23. Xishem

So you got the moles of urea correct?

24. Rogue

Oh, never mind, I realize.

25. anonymous

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)

26. anonymous

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

27. Rogue

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.

28. Rogue

The rest is just a math exercise.

29. Xishem

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.

30. Xishem

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

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