Question

In a sugar mill, sugar cane is shredded, mixed with water and crushed between rollers. The resulting juice contains \(7-11 \%\) sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\). Assume that \(25 \mathrm{~L}\) of collected juice (of density \(1.0 \mathrm{~g} / \mathrm{mL}\) ) yields \(11 \%\) (by mass) sucrose. What is the molality of the sucrose solution after \(33 \%\) (by mass) of the water content of the juice has been removed?

Short answer

Answer: The molality of the sucrose solution after 33% of the water content has been removed is 0.539 mol/kg.

Step-by-step solution

Determine the initial mass of sucrose and water in the sugarcane juice

We are given that 25 L of juice contains 11% (by mass) sucrose. Since the density of the juice is 1.0 g/mL, we can determine the initial mass of juice: Initial_mass_juice = Volume × Density Initial_mass_juice = 25 L × (1000 mL / L) × 1.0 g/mL = 25000 g Now, we can find the initial mass of sucrose and water: Initial_mass_sucrose = 0.11 × Initial_mass_juice Initial_mass_sucrose = 0.11 × 25000 g = 2750 g Initial_mass_water = Initial_mass_juice - Initial_mass_sucrose Initial_mass_water = 25000 g - 2750 g = 22250 g

Calculate the mass of water removed

We are given that 33% (by mass) of the water content is removed. So, let's find the mass of water removed: Mass_water_removed = 0.33 × Initial_mass_water Mass_water_removed = 0.33 × 22250 g = 7342.50 g

Find the final mass of water and the total mass of the solution

Knowing the mass of water removed, we can now determine the final mass of water: Final_mass_water = Initial_mass_water - Mass_water_removed Final_mass_water = 22250 g - 7342.50 g = 14907.50 g The total mass of the solution is the sum of the final mass of water and the mass of sucrose: Total_mass_solution = Final_mass_water + Initial_mass_sucrose Total_mass_solution = 14907.50 g + 2750 g = 17657.50 g

Calculate the molality of the sucrose solution

To find the molality, we first calculate the moles of sucrose: Molar_mass_sucrose = 12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 g/mol Moles_sucrose = Initial_mass_sucrose / Molar_mass_sucrose Moles_sucrose = 2750 g / 342 g/mol = 8.04 mol Now we can calculate the molality of the sucrose solution: Molality = Moles_sucrose / (Final_mass_water / 1000) Molality = 8.04 mol / (14907.50 g / 1000) = 8.04 mol / 14.9075 kg = 0.539 mol/kg The molality of the sucrose solution after 33% of the water content has been removed is 0.539 mol/kg.

Key concepts

Sucrose Solution Concentration

Understanding the concentration of a sucrose solution is essential when working with solutions in chemistry. Molality, a measure of concentration, expresses the number of moles of solute, in this case, sucrose, per kilogram of solvent, which is water. Unlike molarity, which depends on the total volume of the solution, molality is only concerned with the mass of the solvent, making it temperature-independent.

Calculating the molality involves a few steps: first, determining the initial amounts of sucrose and water, then accounting for any changes, such as water removal. Once the final amount of solvent is known, the number of moles of sucrose can be calculated using its molar mass, and finally, these figures can be used to compute the molality using the formula:
\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \].

For our exercise, the steps clearly demonstrated how to arrive at the molality after a specific amount of water was removed from the sugar cane juice. It is crucial for students to grasp the concept of molality since it is widely used in various chemistry applications such as colligative properties and reaction stoichiometry.

Chemistry Problem Solving

Problem-solving in chemistry typically involves a systematic approach that starts with understanding the problem, identifying the relevant data, and then applying chemical principles and mathematical calculations to find a solution. In this case, the problem was to calculate the molality of a sucrose solution after a percentage of water had been removed.

To tackle chemistry problems effectively, it's essential to break them down into manageable steps. By methodically working through each step of the process, as seen in our exercise, students can make complex problems more approachable. This approach also helps in minimizing errors, ensuring units are consistent, and applying conversion factors correctly.

In the provided example, the calculation was simplified into four key steps: finding initial masses, determining the mass of water removed, calculating the new mass of water, and, finally, using these values to find the molality. This segmented approach aids in concept retention and enhances problem-solving skills.

Solution Preparation

Preparing a solution requires a precise understanding of the concentration desired and the components being mixed. The process often involves dissolving a specific amount of solute, such as sucrose, into a solvent, like water. In labs, this involves careful measurement and sometimes the removal or addition of solvent to achieve a target concentration.

In the context of sugar cane juice, preparing a solution with a desired sucrose concentration entails first establishing the initial concentration followed by the adjustment of water content. As the problem shows, after calculating the initial mass of the solvent and solute, we consider any changes to those masses — in our case, the removal of 33% of the water.

The expertise in solution preparation is not just useful for industrial processes such as those at a sugar mill but is also a fundamental skill in chemistry labs. Techniques may range from simple dilutions to more complex procedures requiring multiple steps to achieve a precise concentration, as evidence by our example.