ω (C6H12O6) = 5%
ω (C12H22O11) = 5%
π (C6H12O6)? π (C12H22O11)
1) Assume that the densities of 5% glucose solution and 5% sucrose solution are approximately equal;
2) m = ω * m solution / 100%;
3) n = m / M = (ω * m solution) / (100% * M);
4) Cm = n / V solution = (ω * m solution) / (100% * M * V solution);
5) π = i * Cm * R * T = (i * ω * m solution * R * T) / (100% * M * V solution);
6) π (C6H12O6) / π (C12H22O11) = ((i * ω * m solution * R * T) / (100% * M (C6H12O6) * V solution)) / ((i * ω * m solution * R * T) / (100% * M (C12H22O11) * V solution)) = M (C12H22O11) / M (C6H12O6) = 342/180 = 1.9.
Answer: The osmotic pressure of the C6H12O6 solution is 1.9 times higher.
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