An aqueous solution is 10.% glucose by mass (d = 1.039 g/mL at 20°C). Calculate its freezing point, boiling point at 1 atm, and osmotic pressure

The osmotic pressure is the least amount of pressure that must be applied to a solution in order to block the passage of the solution's pure solvent through a semipermeable membrane.
It can alternatively be described as a measurement of a solution's propensity to osmotically absorb a pure solvent. The highest osmotic pressure that may form in a solution if it were cut off from its pure solvent by a semipermeable membrane is known as the potential osmotic pressure.
When two solutions with various solute concentrations are separated by a selectively permeable membrane, osmosis takes place.
From the low-concentration solution to the solution with increasing solute concentration, solvent molecules move across the membrane preferentially. Up until equilibrium is reached, solvent molecules will continue to move.

we know that

Mass = [tex]volume \times density[/tex]

Mass = [tex]1.039 \times 1000 = 1039[/tex]gm solution

Then the amount of the solute and solvent are

Mass = [tex]0.10 \times 1039 = 103.9 g[/tex] solute

Mass = [tex]0.9 \times 1039 = 935.1g[/tex] solvent

now moles of glucose = [tex]\frac{103.9}{180} = 0.577 moles[/tex]

now molarity M = [tex]\dfrac{moles}{volume} = \dfrac{0.577}{1} = 0.577[/tex]

we know that

[tex]\Delta T = K_b \times m = 0.577 \times 0.617 = 0.32[/tex]

hence the boiling point is

[tex]\Delta T = T_b - T\\T_b = 100 + 0.32 = 100.32[/tex]

osmotic pressure = [tex]\pi = MRT\\\pi = 0.577 \times 0.082 \times 293 =13.9 atm[/tex]

To learn more about osmotic pressure with the given link

https://brainly.com/question/12497098

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