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Chapter 10: Problem 77

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 atm at a temperature of \(22^{\circ} \mathrm{C} ?\) Sucrose is a nonelectrolyte.

Short Answer

Expert verified
To prepare 1.0 L of the aqueous sucrose solution having an osmotic pressure of 15 atm at a temperature of 22°C, first calculate the molarity (M) using the formula: Osmotic Pressure = Molarity × Gas Constant × Temperature. Find the molarity to be 0.685 mol/L. Next, calculate the mass of sucrose needed, which is approximately 234.27 g. Finally, dissolve 234.27 g of sucrose in water and adjust the volume to 1.0 L.

Step by step solution

01

Identify the given information

We are given: - Osmotic pressure = 15 atm - Temperature = 22°C (which we will convert to Kelvin) - Molecular formula of sucrose = C12H22O11
02

Convert the temperature to Kelvin

To convert the temperature from Celsius to Kelvin, add 273.15: Temperature = 22 + 273.15 = 295.15 K
03

Determine the osmotic pressure formula and solve for molarity

We will use the formula: Osmotic Pressure = Molarity × Gas Constant × Temperature Where the gas constant (R) has a value of 0.08206 L atm/mol K. Now we will solve for Molarity (M): 15 atm = M × 0.08206 L atm/mol K × 295.15 K M = (15 atm) / (0.08206 L atm/mol K × 295.15 K) M = 0.685 mol/L
04

Calculate the mass of sucrose needed

To prepare 1.0 L of the solution, we need to determine the mass of sucrose needed for a molarity of 0.685 mol/L. First, calculate the molar mass of sucrose: Molar mass of sucrose = (12 × 12) + (22 × 1) + (11 × 16) = 342 g/mol Now, calculate the mass of sucrose needed: Mass of sucrose = Molarity × Volume × Molar mass of sucrose Mass of sucrose = 0.685 mol/L × 1.0 L × 342 g/mol Mass of sucrose ≈ 234.27 g
05

Prepare the solution

To prepare 1.0 L of the aqueous sucrose solution having an osmotic pressure of 15 atm at a temperature of 22°C, dissolve approximately 234.27 g of sucrose in water and adjust the volume to 1.0 L.

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Most popular questions from this chapter

Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound. A \(0.378-m\) solution is prepared by dissolving 38.4 g sodium diatrizoate (NaDTZ) in \(1.60 \times 10^{2} \mathrm{mL}\) water at \(31.2^{\circ} \mathrm{C}\) (the density of water at \(31.2^{\circ} \mathrm{C}\) is \(0.995 \mathrm{g} / \mathrm{cm}^{3}\) ). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at \(31.2^{\circ} \mathrm{C}\) is 34.1 torr?

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A 0.500 -g sample of a compound is dissolved in enough water to form \(100.0 \mathrm{mL}\) of solution. This solution has an osmotic pressure of 2.50 atm at \(25^{\circ} \mathrm{C}\). If each molecule of the solute dissociates into two particles (in this solvent), what is the molar mass of this solute?

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