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?
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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!

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.

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

Thanks a whole lot =)

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.

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

This has been very helpful!!!!

It looks like you're on the right track, but there seems to be a slight mistake in your equations. Let's break down the problem and go through the steps again.

Given:
Mass of the sample (glucose + sucrose) = 1.10g
Total solution volume = 25.0mL = 0.025L
Osmotic pressure of the solution = 3.78atm
Temperature = 298K

First, let's calculate the molarity (M) using the formula for osmotic pressure:

pi = MRT

Given:
pi = 3.78atm
R = 0.0821 L·atm/(mol·K)
T = 298K

Substituting these values into the formula, we can solve for M (molarity):

3.78atm = M * (0.0821 L·atm/(mol·K)) * 298K

Now, you can calculate M.

Next, we can use the molarity (M) and the volume (0.025L) to calculate the moles (mol) of the solute in the solution using the formula:

moles = M * volume

Now that we have the moles, we can set up the two equations to solve for the mass percentage of glucose and sucrose.

Let's use X to represent the mass of glucose and Y to represent the mass of sucrose.

The total mass is given as 1.10g, so we can write the equation:

X + Y = 1.10g

Now, let's consider the moles of each substance. The moles of glucose divided by the molar mass of glucose plus the moles of sucrose divided by the molar mass of sucrose should be equal to the total moles we calculated earlier:

(X/molar mass of glucose) + (Y/molar mass of sucrose) = moles

Now, let's substitute the values into the equation. The molar mass of glucose is 180.162g/mol and the molar mass of sucrose is 342.308g/mol.

Here's the equation:

(X/180.162g/mol) + (Y/342.308g/mol) = moles

Now, you can solve these two equations simultaneously to find the values of X and Y, which represent the masses of glucose and sucrose, respectively.

Once you have the values for X and Y, you can calculate the mass percent of each substance by dividing each mass by the total mass (1.10g) and multiplying by 100.

I hope this helps! Let me know if you have any more questions or if there's anything else I can assist you with.

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 = ??