Question

A sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) solution that is \(45.0 \%\) sucrose by mass has a density of \(1.203 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Calculate its (a) molarity. (b) molality. (c) vapor pressure \(\left(\mathrm{vp} \mathrm{H}_{2} \mathrm{O}\right.\) at \(\left.25^{\circ} \mathrm{C}=23.76 \mathrm{~mm} \mathrm{Hg}\right)\). (d) normal boiling point.

Short answer

2. What is the molality of the sucrose solution? 3. What is the vapor pressure of the sucrose solution? 4. What is the normal boiling point of the sucrose solution?

Step-by-step solution

Calculate mass of sucrose and water used in the solution.

Since the solution is 45.0% sucrose by mass, this means that in a 100 g solution, there would be 45 g of sucrose, and 55 g of water.

Calculate moles of sucrose and water.

To determine molarity and molality, we need to find the moles of sucrose and water in our 100 g solution. Using the molar mass, divide the mass of each component by their molar mass: Moles of sucrose = \(\frac{45 \,g}{342.3 \,g/mol} = 0.1314 \,mol\) Moles of water = \(\frac{55 \,g}{18.02 \,g/mol} = 3.052 \,mol\)

Calculate the volume of the solution.

We also need the volume of the solution for molarity calculations. Using the density provided, divide the mass of the 100 g solution by the given density: Volume = \(\frac{100 \,g}{1.203 \,g/mL} = 83.13 \,mL\) or \(0.08313 \,L\)

Calculate molarity.

Molarity is defined as moles of solute per liter of solution. Now that we know the moles of sucrose and the volume of the solution, divide the moles of sucrose by the volume: Molarity = \(\frac{0.1314\,mol}{0.08313\,L} = 1.58\, M\)

Calculate molality.

Molality is defined as moles of solute per kilogram of solvent. Divide the moles of sucrose by the mass of water in kilograms: Molality = \(\frac{0.1314\,mol}{0.055\,kg} = 2.39\, m\)

Calculate vapor pressure.

According to Raoult's Law, the vapor pressure of a solution is the product of the mol fraction of the solvent and the vapor pressure of the pure solvent. First, calculate the mol fraction of water in the solution: Mol fraction of water = \(\frac{3.052\,mol}{3.052\,mol + 0.1314\,mol} = 0.9587\) Now, multiply the mol fraction of water with the vapor pressure of pure water: Vapor pressure of the solution = \(0.9587 * 23.76\,mm\,Hg = 22.77\,mm\,Hg\)

Calculate the normal boiling point.

Finally, to find the normal boiling point, determine the boiling point elevation using the formula: Boiling point elevation = \(\Delta T_{b} = K_{b} * m\) For water, the boiling point elevation constant, \(K_{b} = 0.512\,K \cdot kg/mol\) Now substitute the molality and equilibrium constant values: \(\Delta T_{b} = 0.512\,K \cdot kg/mol * 2.39\, mol/kg = 1.225 \,K\) The normal boiling point of water is \(100^{\circ}C\); thus, the boiling point of the sucrose solution is: Normal boiling point = \(100^{\circ}C + 1.225^{\circ}C = 101.225^{\circ}C\) In summary, the sucrose solution has the following properties: (a) Molarity = \(1.58\, M\) (b) Molality = \(2.39\, m\) (c) Vapor pressure = \(22.77\,mm\,Hg\) (d) Normal boiling point = \(101.225^{\circ}C\)