Problem 67 - General Chemistry

Question

A sample of \(0.0512\) mol of iron(III) chloride, \(\mathrm{FeCl}_{3}\), was dissolved in water to give \(25.0 \mathrm{~mL}\) of solution. What is the molarity of the solution?

Step-by-Step Solution

Answer

The molarity of the solution is 2.048 M.

Step 1: Understand the Problem

We are given a sample of iron(III) chloride (\(\mathrm{FeCl}_{3}\)) with a known amount in moles and the final volume of the solution. Our objective is to find its molarity.

Step 2: Review Molarity Formula

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 \(M\) is the molarity, \(n\) is the number of moles, and \(V\) is the volume of the solution in liters.

Step 3: Convert Volume to Liters

The given volume is \(25.0\, \mathrm{mL}\). We need to convert this to liters by using the conversion factor:\[1 \text{ L} = 1000 \text{ mL}\]Thus,\[25.0 \text{ mL} = \frac{25.0}{1000} = 0.025 \text{ L}\]

Step 4: Calculate Molarity

Using the formula for molarity,\[M = \frac{n}{V}\]where \(n = 0.0512\, \text{mol}\) and \(V = 0.025\, \text{L}\), substitute the values:\[M = \frac{0.0512}{0.025} = 2.048\, \text{M}\]

Step 5: State the Final Answer

The molarity of the \(\mathrm{FeCl}_{3}\) solution is \(2.048\, \text{M}\).

Key Concepts

Molarity Formula

In chemistry, molarity plays a crucial role in characterizing solutions. It helps in understanding the concentration of a solution, which is expressed as the number of moles of a solute per liter of solution. This is essential for various calculations and reactions. The formula for molarity is given by:

• \[ M = \frac{n}{V} \]

where \( M \) represents molarity, \( n \) signifies the number of moles of the solute, and \( V \) is the volume of the solution in liters. This equation shows that molarity is inversely proportional to volume and directly proportional to the number of moles. A higher number of moles or a smaller volume will result in a higher molarity, making this formula a fundamental tool in the quantitative analysis of solutions.

Unit Conversion

Unit conversion is vital when calculating molarity because the volume must be converted into liters. This is a common step that can sometimes trip people up. To achieve this, one must remember that:

• 1 liter equals 1000 milliliters.

Knowing this, you can convert any given volume in milliliters to liters by dividing by 1000. In the problem given, the volume was 25.0 mL. By converting it using the conversion factor, you get:

• \[ 25.0 \, \text{mL} = \frac{25.0}{1000} = 0.025 \, \text{L} \]

This allows for the accurate application of the molarity formula by ensuring the volume is in the correct units. Proper unit conversion ensures your calculations are accurate and meaningful.

Iron(III) Chloride Solution

Iron(III) chloride, often denoted as \( \text{FeCl}_3 \), is a chemical compound that can dissolve in water to form a solution. When it dissolves, it dissociates into iron and chloride ions. This dissolution is crucial because it determines the number of moles present in the solution. In the given exercise, you've been provided with \( 0.0512 \) moles of \( \text{FeCl}_3 \). By knowing the moles and understanding how the substance interacts in water, you can calculate properties like molarity or predict other chemical behaviors. Solutions like these are frequently used in laboratories and various industrial processes, making the understanding of their properties critical for solving many chemistry problems.

Moles and Volume Calculation

Calculating moles and volume is a foundational part of determining molarity. In this exercise, you were given the number of moles directly, as \( 0.0512 \) moles of \( \text{FeCl}_3 \). However, in other scenarios, you might derive moles from mass using the formula:

• \[ n = \frac{m}{M_w} \]

where \( n \) is the number of moles, \( m \) is the mass of the substance, and \( M_w \) is the molar mass.Once you have moles and volume—converted into liters as shown—the molarity can be calculated. Remember, precision in these calculations is key, as small errors in determining either moles or volume can lead to significant deviations in molarity, impacting experimental outcomes or further calculations.