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college chem

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A 1.10g- sample contains only glucose and sucrose . When the sample is dissolved in water to a total solution volume of 25.0ML , the osmotic pressure of the solution is 3.78atm at 298K

a.What is the percent by mass of glucose in the sample?

b.What is the percent by mass of sucrose in the sample?
pi = osmotic pressure = MRT
You know pi, R, and T, solve for M
Molarity = moles x L
You know L, solve for moles.
Here are the two equations.
X + Y = 1.10
(X/molar mass glucose) + (Y/molar mass sucrose) = moles.
Solve for X and Y, then you can determine percent of each.
Post your work if you get stuck.

im so lost on the same problem.
ok here is what i have done..

so mols for .025L is 2.9369
And you when i used the two equations i got lost.


and x/(180.162g/mol)+y/(342.308g/mol)=2.9362

i think i did this wrong. Please help!

  • college chem - ,

    How in the world did you obtain 117 for the molarity? Your math skills need improving.
    pi = MRT so
    M = pi/RT = 3.78/0.08206*298 = 0.1545 (you are allowed only 3 s.f. in the answer but I always carry one more place, then round at the end).
    Then M = moles/L or moles = M x L = 0.1545 x 0.025 = 0.00386.
    So the equations are
    X + Y = 1.10
    (X/180.2) + (Y/342.3) = 0.00386
    unless I've goofed somewhere. So we solve the first equation for X.
    X = 1.10 - Y and substitute into equation 2.
    (1.10-Y/180.2) + (Y/342.3) = 0.00386 which leaves you with just one unknown, Y, for which to solve. That should give you the grams of Y and I believe we let Y = grams sucrose. Solve for Y, then plug Y into X+Y= 1.10 to obtain X.
    %glucose = (X/1.10)*100 = ??
    %sucrose = (Y/1.10)*100 = ??

  • college chem - ,

    haha..i thought i needed to use torr instead of atm that's how i got that M

    Thanks a whole lot =)

  • college chem - ,

    so i went through the process that DrBob put up and i got .662g for y and .438g for x and got basically 39.8% for the glucose and 60.2% for sucrose and it said i was wrong. If anybody can do the math and double check let me know.

    I basically started off with finding a common denominator, after substituting 1.10-y for x, and went from there. Let me know what you get.

  • college chem - ,

    I worked through it and obtained
    0.248 g for glucose (X) and 0.852 g for sucrose (Y) which makes the percents
    glucose = 22.54 which rounds to 22.5%
    sucrose = 77.45 which rounds to 77.4%.
    I plugged both masses back into the equations and those values satisfied both equations; therefore, the math should be ok. The chemistry looks ok to me, too.

  • college chem - ,

    great thank you!! ill have to check over mine then!!

  • college chem - ,

    This has been very helpful!!!!

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