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March 6, 2015

March 6, 2015

Posted by **mm** on Monday, January 25, 2010 at 10:27pm.

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..

M=117.4782mol/L

so mols for .025L is 2.9369

And you when i used the two equations i got lost.

x=1.10g-y

y=1.10g-x

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

i think i did this wrong. Please help!

- college chem -
**DrBob222**, Monday, January 25, 2010 at 10:55pmHow 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 -
**mm**, Monday, January 25, 2010 at 11:20pmhaha..i thought i needed to use torr instead of atm that's how i got that M

Thanks a whole lot =)

- college chem -
**Taylor G**, Monday, January 25, 2010 at 11:34pmso 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 -
**DrBob222**, Monday, January 25, 2010 at 11:58pmI 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 -
**Taylor G**, Tuesday, January 26, 2010 at 12:07amgreat thank you!! ill have to check over mine then!!

- college chem -
**Scott**, Monday, July 26, 2010 at 12:18amThis has been very helpful!!!!

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