Problem 54
Question
A \(23.2 \%\) by weight aqueous solution of sucrose has a density of \(1.127 \mathrm{~g} / \mathrm{mL}\). Calculate the molarity of sucrose in this solution.
Step-by-Step Solution
Verified Answer
The molarity of the sucrose solution is approximately 0.764 M.
1Step 1: Understand the Definition of Molarity
Molarity is defined as the number of moles of solute per liter of solution. It is represented by the formula \( M = \frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume in liters.
2Step 2: Calculate the Mass of Sucrose in 1 L of Solution
Assume we have 1000 mL (or 1 L) of solution. Since the density of the solution is \(1.127 \text{ g/mL}\), the total mass of 1 L of solution is \(1.127 \times 1000 = 1127 \text{ g}\).
3Step 3: Calculate the Mass of Sucrose
Given the solution is \(23.2\%\) sucrose by weight, the mass of sucrose in 1127 g of solution is \(0.232 \times 1127 = 261.464 \text{ g}\).
4Step 4: Determine the Molar Mass of Sucrose
The chemical formula for sucrose is \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \). Calculating its molar mass: \[(12 \times 12.01) + (22 \times 1.01) + (11 \times 16.00) = 342.30 \text{ g/mol}\].
5Step 5: Calculate the Moles of Sucrose
Use the molar mass to convert mass to moles: \[ n = \frac{261.464}{342.30} = 0.7639 \text{ moles}\].
6Step 6: Calculate the Molarity of Sucrose
Molarity \( M \) is moles of solute per liter of solution. \[ M = \frac{0.7639 \text{ moles}}{1 \text{ L}} = 0.7639 \text{ M} \].
Key Concepts
Molar MassDensityWeight PercentConcentration
Molar Mass
Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). To calculate the molar mass, you need the molecular formula of the compound. For example, for sucrose with the formula \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \), you need to consider the atomic masses of carbon (C), hydrogen (H), and oxygen (O).
- Carbon has an atomic mass of approximately 12.01 g/mol.
- Hydrogen has an atomic mass of about 1.01 g/mol.
- Oxygen has an atomic mass of roughly 16.00 g/mol.
Density
Density is a measure of how much mass is contained in a given volume. It is usually expressed in units like grams per milliliter (g/mL) or kilograms per liter (kg/L). Understanding density is crucial for converting between mass and volume, as it directly relates these two properties.In the context of solutions, knowing the density allows us to calculate the total mass of a given volume of solution. This is particularly useful when a problem requires determining how much solute is present in a specific volume of solution. For example, given a solution with a density of 1.127 g/mL, the mass of 1 liter (1000 mL) of this solution can be calculated as:\[ 1.127 \times 1000 = 1127 \text{ g}. \]This relationship allows further calculations involving mass and moles, leading to finding the molarity.
Weight Percent
Weight percent (also known as mass percent) is a way of expressing the concentration of a solute in a solution. It describes the mass of the solute divided by the total mass of the solution, multiplied by 100 to give a percentage.For instance, a 23.2% weight percent sucrose solution means that there are 23.2 grams of sucrose in every 100 grams of solution. To calculate the mass of sucrose in any amount of solution, multiply the weight percent by the total mass of the solution and divide by 100.In our given problem, for a solution weighing 1127 grams:\[ 0.232 \times 1127 = 261.464 \text{ grams of sucrose}. \]This value can then be used in further calculations to determine molarity.
Concentration
Concentration refers to the quantity of solute present in a given quantity of solvent or solution. There are several ways to express concentration, with molarity being one of the most common in chemistry.Molarity \( M \) is defined as the number of moles of solute per liter of solution. To convert from mass to moles, you use the molar mass of the solute. Thus, concentration calculations often involve:
- Determining the mass of solute and converting it to moles using molar mass.
- Dividing moles of solute by the volume of solution in liters.
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