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Chapter 11: Problem 75

How would you prepare \(1.0 \mathrm{~L}\) of an aqueous solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) having an osmotic pressure of \(15 \mathrm{~atm}\) at a temperature of \(22^{\circ} \mathrm{C} ?\) Sucrose is a nonelectrolyte.

Short Answer

Expert verified
To prepare 1.0 L of an aqueous solution of sucrose with an osmotic pressure of 15 atm at 22°C, first, calculate the molar concentration using the osmotic pressure formula, which gives \(c ≈ 0.6595 \mathrm{ ~M}\). Then, calculate the mass of sucrose required by multiplying the molar concentration, molar mass of sucrose, and the volume of the solution, resulting in \(m ≈ 225.63 \mathrm{ ~g}\). Finally, dissolve approximately 225.63 g of sucrose in water and adjust the total volume to be 1.0 L.

Step by step solution

01

1. Convert temperature to Kelvin

To convert the given temperature from Celsius to Kelvin, add 273.15: $$T(K) = T(^\circ C) + 273.15$$ $$T = 22 ^\circ C + 273.15 = 295.15 K$$
02

2. Calculate molar concentration from the osmotic pressure formula

Using the osmotic pressure formula, \(\Pi = cRT\), we can isolate the molar concentration (c) and solve for it: $$c = \frac{\Pi}{RT}$$ Plug in the values: $$c = \frac{15 \mathrm{~atm}}{(0.0821 \mathrm{ ~L\cdot atm/mol \cdot K})(295.15 \mathrm{ ~K})}$$ Calculate the molar concentration (c): $$c ≈ 0.6595 \mathrm{ ~M}$$
03

3. Calculate the mass of sucrose needed

To calculate the mass of sucrose needed, multiply the molar concentration by the molar mass of sucrose and the volume of the solution: Molar mass of sucrose (C12H22O11) = 12(12.01) + 22(1.008) + 11(16.00) = 342.3 g/mol Mass of sucrose needed (m) = Molar concentration (c) × Molar mass × Volume (V): $$m = (0.6595 \mathrm{ ~M})(342.3 \mathrm{ ~g/mol})(1.0 \mathrm{ ~L})$$ $$m ≈ 225.63 \mathrm{ ~g}$$
04

4. Prepare the solution

To prepare 1.0 L of the aqueous sucrose solution with osmotic pressure of 15 atm at 22°C, dissolve approximately 225.63 g of sucrose in water, and adjust the total volume to be 1.0 L.

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