What mass of sucrose (C12H22O11) should be combined with 477g of water to make a solution with an osmotic pressure of 8.15atm at 275K ? (Assume the density of the solution to be equal to the density of the solvent.)

Answer 1

pi = MRT
Substitute and solve for M.
Now you must convert M to mols sucrose needed.
M = mols sucrose/L solution.
mols = g/molar mass = g/342 so
M = g sucrose/342

Now calculate the volume in L.
You will have 447g solvent + g sucrose as a mass. The density is 1.00 g/mL so the volume in mL will be 447+g sucrose/1 and convert that to L will be 447g H2O + g sucrose/1000 and substitute all of that into
0.36M = (g/342)/(447+g/1000) and solve for g which is g sucrose. I obtained approx 63g but you need to confirm all of these numbers. Just assuming 63g sucrose is close let's calculate pi.
mols sucrose = 63/342 = 0.184
g solution = 447 + 63 = 510 g and with a density of 1.00 g/mL that will be 510 mL or 0.510 L so M = 0.184/0.510 = 0.36M
Then pi = 0.36*0.08205*275 = 8.12 atm so that 63g sucrose must be close.