Answer:
[tex]\large \boxed{79 \, \%}[/tex]
Explanation:
I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.
We have two conditions:
(1) Mass of glucose + mass of sucrose = 1.10 g
(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm
Let g = mass of glucose
and s = mass of sucrose. Then
g/180.16 = moles of glucose, and
s/342.30 = moles of sucrose. Also,
g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and
s/(342.30×2.50) = s/855.8 = molar concentration of sucrose.
1. Set up the osmotic pressure condition
Π = cRT, so
[tex]\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}[/tex]
Now we can write the two simultaneous equations and solve for the masses.
2. Calculate the masses
[tex]\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}[/tex]
We have 0.229 g of glucose and 0.871 g of sucrose.
3. Calculate the mass percent of sucrose
[tex]\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}[/tex]
