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What is the osmotic pressure of a solution made by dissolving 75.0 g of glucose, C₆H₁₂O₆, in enough water to form 300.0 ml of solution at 7.00 ∘C?

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Final answer:

The osmotic pressure of the glucose solution at 7.00 °C is calculated to be 31.78 atm, using the formula Π = iMRT with the molarity determined by the mass of glucose, its molar mass, and the volume of the solution.

Step-by-step explanation:

Calculation of the Osmotic Pressure of a Glucose Solution

The osmotic pressure of a solution can be calculated using the formula: Π = iMRT, where Π is the osmotic pressure, i is the van't Hoff factor (which is 1 for glucose as it doesn't dissociate into multiple particles), M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin.

To find the molarity (M), we use the mass of glucose (75.0 g), the molar mass of glucose (C₆H₁₂O₆, which is approximately 180.16 g/mol), and the volume of the solution (300.0 ml or 0.300 L).

M = mass of solute (g) / (molar mass of solute (g/mol) × volume of solution (L))

M = 75.0 g / (180.16 g/mol × 0.300 L) = 1.39 M

Convert the temperature from Celsius to Kelvin: T(K) = 7.00 °C + 273.15 = 280.15 K

Using the ideal gas constant R = 0.0821 L·atm/mol·K, the osmotic pressure is calculated as follows:

Π = (1) × (1.39 M) × (0.0821 L·atm/mol·K) × (280.15 K)

Π = 31.78 atm

Therefore, the osmotic pressure of the glucose solution at 7.00 °C is 31.78 atm.

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