Question

You make 1.000 L of an aqueous solution that contains 35.0 \(\mathrm{g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) .\) (a) What is the molarity of sucrose in this solution? (b) How many liters of water would you have to add to this solution to reduce the molarity you calculated in part (a) by a factor of two?

Short answer

(a) The molarity of sucrose in the initial solution is \(0.1022 M\). (b) To reduce the molarity by a factor of two, \(1.000 L\) of water should be added to the solution.

Step-by-step solution

Find the molar mass of sucrose

To find the molar mass of sucrose (\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)), we need to sum the molar masses of all the individual atoms in a molecule of sucrose. Sucrose molecule has: - 12 carbon atoms (C) with molar mass 12.01 g/mol - 22 hydrogen atoms (H) with molar mass 1.01 g/mol - 11 oxygen atoms (O) with molar mass 16.00 g/mol Molar mass of sucrose = (12 × 12.01) + (22 × 1.01) + (11 × 16.00) = 144.12 + 22.22 + 176 = 342.34 g/mol.

Calculate the moles of sucrose

Now, we have the mass of sucrose (35.0 g) and the molar mass of sucrose (342.34 g/mol). We can find the number of moles using the formula: Moles of sucrose = mass of sucrose / molar mass of sucrose Moles of sucrose = 35.0 g / 342.34 g/mol = 0.1022 mol

Calculate the initial molarity of the solution

Using the moles of sucrose and the volume of the solution, we can find the molarity with the following formula: Molarity = moles of solute / volume of solution in liters Molarity = 0.1022 mol / 1.000 L = 0.1022 M So, the molarity of sucrose in the initial solution is 0.1022 M.

Calculate the volume of water to add to halve the molarity

Now we want to halve the initial molarity, which means we need to find a new volume of solution that would make the molarity 0.0511 M (half of 0.1022 M). We can use the formula: Molarity_new × Volume_new = Molarity_initial × Volume_initial Volume_new = Molarity_initial × Volume_initial / Molarity_new Volume_new = (0.1022 M × 1.000 L) / 0.0511 M = 2.000 L So, we need to have 2.000 L of solution to halve the initial molarity. Since we started with 1.000 L of solution, we need to add an additional 1.000 L of water to the solution. Answer: (a) The molarity of sucrose in the initial solution is 0.1022 M. (b) To reduce the molarity by a factor of two, 1.000 L of water should be added to the solution.

Key concepts

Understanding Sucrose

Sucrose is a type of sugar composed of two simpler sugars, glucose and fructose. It is commonly found in various sweet foods, like table sugar, and is used in many chemical experiments. When you're analyzing sucrose in a solution, you are often dealing with its chemical formula, which is \( \text{C}_{12} \text{H}_{22} \text{O}_{11} \). This formula tells us about the types and numbers of atoms present in one molecule of sucrose.
- **Carbon (C):** 12 atoms per molecule- **Hydrogen (H):** 22 atoms per molecule- **Oxygen (O):** 11 atoms per molecule

Knowing the chemical structure is crucial because it affects how sucrose interacts in solutions, and influences calculations like molarity, which refers to the concentration of a solution. It’s important to remember that concentration plays a critical role in reactions involving sucrose, such as fermentation or hydrolysis.

Calculating Molar Mass

Molar mass is a key concept in chemistry and crucial for calculating the molarity of solutions. The molar mass refers to the mass of one mole of a given substance, expressed in grams per mole (g/mol). For sucrose, its molar mass can be calculated by summing the atomic masses of all the atoms in its chemical formula. Here's a breakdown:

• Each carbon atom (C) has an atomic mass of 12.01 g/mol, and with 12 carbon atoms, the total mass from carbon is \( 12 \times 12.01 = 144.12 \text{ g/mol} \).
• Each hydrogen atom (H) has an atomic mass of 1.01 g/mol, contributing \( 22 \times 1.01 = 22.22 \text{ g/mol} \).
• Each oxygen atom (O) weighs 16.00 g/mol, adding \( 11 \times 16.00 = 176.00 \text{ g/mol} \) to the total.

Adding these together gives a molar mass for sucrose of 342.34 g/mol. Knowing the molar mass allows you to convert between grams and moles, a necessary step for determining the molarity of a sucrose solution.

Exploring Dilution

Dilution is a process in chemistry where you reduce the concentration of a solute in a solution, usually by adding more solvent, such as water. It is a commonly applied concept when one needs to prepare solutions of varying concentrations. Here's how it works with sucrose.
Let's say you have a solution with a specific molarity, which expresses concentration. To dilute it, you increase the volume of the solution while keeping the amount of solute (sucrose) constant. This causes the molarity to decrease because the same number of moles is now dissolved in a greater volume.
The calculation for dilution often uses the formula:\[ \text{Molarity}_{\text{new}} \times \text{Volume}_{\text{new}} = \text{Molarity}_{\text{initial}} \times \text{Volume}_{\text{initial}} \]This equation helps us determine how much solvent to add. In the exercise above, to halve the sucrose solution's molarity, we determine the final volume must be twice the initial one. Thus, if you start with 1.000 L, you add enough solvent to reach 2.000 L.