Problem 60 - Chemistry: The Central Science in SI Units, Global Edition

Question

You make \(1.000 \mathrm{~L}\) 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?

Step-by-Step Solution

Answer

The molarity of the sucrose solution is approximately 0.102 M. To reduce the molarity by a factor of two, 1.000 L of water needs to be added to the solution.

Step 1: Part (a): Finding the Molarity of the Sucrose Solution

To find the molarity of the sucrose solution, we need to calculate the number of moles of sucrose in the solution. We can do this using the given mass of sucrose and its molar mass. After that, we will use the molarity formula to calculate the molarity. 1. Calculate the molar mass of sucrose: The molar mass of sucrose (\(C_{12}H_{22}O_{11}\)) can be calculated by adding the molar masses of its individual atoms: Molar mass of sucrose = (12 × molar mass of C) + (22 × molar mass of H) + (11 × molar mass of O) Molar mass of sucrose = (12 × 12.01 g/mol) + (22 × 1.008 g/mol) + (11 × 16.00 g/mol) ≈ 342.30 g/mol 2. Calculate the number of moles of sucrose: Number of moles = (mass of sucrose) / (molar mass of sucrose) Number of moles = (35.0 g) / (342.30 g/mol) ≈ 0.102 mol 3. Calculate the molarity of the sucrose solution: Molarity = (number of moles of solute) / (volume of solution in liters) Molarity ≈ 0.102 mol / 1.000 L ≈ 0.102 M So, the molarity of the sucrose solution is approximately 0.102 M.

Step 2: Part (b): Finding the Volume of Water Needed to Dilute the Solution

Now, we need to find the volume of water that needs to be added to the solution in order to reduce the molarity by a factor of two. To do this, we will use the dilution formula: \(C_1 V_1 = C_2 V_2\) Where: - \(C_1\) is the initial concentration (0.102 M) - \(V_1\) is the initial volume (1.000 L) - \(C_2\) is the final concentration (half of the initial concentration, or 0.051 M) - \(V_2\) is the final volume (unknown) 1. Rearrange the dilution formula to solve for \(V_2\): \(V_2 = \frac{C_1 V_1}{C_2}\) 2. Insert the values and calculate \(V_2\): \(V_2 = \frac{0.102 M * 1.000 L}{0.051 M} ≈ 2.000 L\) 3. Calculate the volume of water to be added: Volume of water to be added = final volume - initial volume Volume of water to be added = 2.000 L - 1.000 L = 1.000 L So, 1.000 L of water needs to be added to the solution in order to reduce the molarity of sucrose by a factor of two.

Key Concepts

Dilution

Dilution is a simple yet powerful concept in chemistry. It involves reducing the concentration of a solute in a solution, often by adding more solvent, such as water.
In the context of solutions, concentration refers to how much solute is present in a given amount of solution. For sucrose solutions, as in the exercise, concentration is commonly measured in molarity. A solution with higher molarity has more sucrose per unit volume than a more diluted solution.
When diluting solutions, we use the dilution formula:

• \( C_1 V_1 = C_2 V_2 \)

Here, \( C_1 \) and \( C_2 \) are the initial and final concentrations, respectively, while \( V_1 \) and \( V_2 \) are the initial and final volumes.
We rearrange the formula to solve for the unknown, often the desired final volume \( V_2 \).
In this exercise, we found that adding 1.000 L of water doubles the original solution's volume, thus halving the concentration of sucrose.

Molar Mass

Molar mass is essential for converting between the mass of a substance and the number of moles. It is the mass of one mole of a given substance, typically expressed in grams per mole (g/mol).
To find the molar mass of a compound, like sucrose (\( C_{12}H_{22}O_{11} \)), sum up the molar masses of all the atoms in the molecule.

• Molar mass of Carbon (C) = about 12.01 g/mol
• Molar mass of Hydrogen (H) = about 1.008 g/mol
• Molar mass of Oxygen (O) = about 16.00 g/mol

Hence, the molar mass calculation for sucrose is:

• \( 12(12.01) + 22(1.008) + 11(16.00) \approx 342.30 \) g/mol

Knowing the molar mass allows you to convert between grams of sucrose and moles of sucrose, a critical step in calculating molarity.

Sucrose Solution

Sucrose, commonly known as table sugar, is a simple sugar with the formula \( C_{12}H_{22}O_{11} \). In solutions, it acts as a solute when dissolved in water. This sucrose solution is described by its molarity, which is the number of moles of sucrose per liter of solution.
To calculate the molarity, start by determining how many moles of sucrose are present. Use the formula:

• Number of moles = \( \frac{\text{mass of sucrose}}{\text{molar mass of sucrose}} \)

For example, if you dissolve 35.0 g of sucrose in water, you can calculate the moles using its molar mass (about 342.30 g/mol).
This results in approximately 0.102 moles.
The solution's molarity is then found by dividing moles by the volume of the solution (in liters), leading to a molarity of approximately 0.102 M.
Such knowledge allows chemists to precisely create solutions with desired concentrations.