Problem 12 - Chemistry: An Atoms-Focused Approach

Question

Calculate the molarity of each of the following solutions: a. 0.150 mol urea \(\left[\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right]\) in \(250.0 \mathrm{mL}\) of solution b. \(1.46 \mathrm{mol} \mathrm{NH}_{4} \mathrm{CH}_{3} \mathrm{CO}_{2}\) in \(1.000 \mathrm{L}\) of solution c. 1.94 mol methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in 5.000 L of solution d. 0.045 mol sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(50.0 \mathrm{mL}\) of solution

Step-by-Step Solution

Answer

Question: Calculate the molarities of the following solutions: a) 0.150 mol of urea (NH2CONH2) in 250.0 mL of solution. b) 1.46 mol of NH4CH3CO2 in a 1.000 L of solution. c) 1.94 mol of methanol (CH3OH) in 5.000 L of solution. d) 0.045 mol of sucrose (C12H22O11) in 50.0 mL of solution. Answer: a) The molarity of the urea solution is \(0.600 \: \frac{\text{mol}}{\text{L}}\). b) The molarity of the \(\mathrm{NH}_{4} \mathrm{CH}_{3} \mathrm{CO}_{2}\) solution is \(1.46 \: \frac{\text{mol}}{\text{L}}\). c) The molarity of the methanol solution is \(0.388 \: \frac{\text{mol}}{\text{L}}\). d) The molarity of the sucrose solution is \(0.900 \: \frac{\text{mol}}{\text{L}}\).

Step 1: a. Molarity of the urea solution

First, convert the given volume into liters: \(250.0 \: \text{mL} = \frac{250.0}{1000} \: \text{L} = 0.250 \: \text{L}\) Now, apply the molarity formula: Molarity = \(\frac{0.150 \: \text{mol}}{0.250 \: \text{L}} = 0.600 \: \frac{\text{mol}}{\text{L}}\) The molarity of the urea solution is \(0.600 \: \frac{\text{mol}}{\text{L}}\).

Step 2: b. Molarity of the \(\mathrm{NH}_{4} \mathrm{CH}_{3} \mathrm{CO}_{2}\) solution

In this case, the volume is already given in liters. Therefore, we can directly apply the molarity formula: Molarity = \(\frac{1.46 \: \text{mol}}{1.000 \: \text{L}} = 1.46 \: \frac{\text{mol}}{\text{L}}\) The molarity of the \(\mathrm{NH}_{4} \mathrm{CH}_{3} \mathrm{CO}_{2}\) solution is \(1.46 \: \frac{\text{mol}}{\text{L}}\).

Step 3: c. Molarity of the methanol solution

As the volume is given in liters, we can directly apply the molarity formula: Molarity = \(\frac{1.94 \: \text{mol}}{5.000 \: \text{L}} = 0.388 \: \frac{\text{mol}}{\text{L}}\) The molarity of the methanol solution is \(0.388 \: \frac{\text{mol}}{\text{L}}\).

Step 4: d. Molarity of the sucrose solution

Convert the given volume into liters: \(50.0 \: \text{mL} = \frac{50.0}{1000} \: \text{L} = 0.050 \: \text{L}\) Now, apply the molarity formula: Molarity = \(\frac{0.045 \: \text{mol}}{0.050 \: \text{L}} = 0.900 \: \frac{\text{mol}}{\text{L}}\) The molarity of the sucrose solution is \(0.900 \: \frac{\text{mol}}{\text{L}}\).

Key Concepts

Solution Concentration

Understanding solution concentration is essential in chemistry. Specifically, molarity is a popular way to express it. Molarity, denoted by a capital M, is defined as the number of moles of a solute per liter of solution. It's expressed in units of moles per liter (mol/L).

This measure helps scientists and students gauge how much of a substance is mixed into a liquid. Common terms that relate to solution concentration include:

• Solute: the substance being dissolved
• Solvent: the liquid that dissolves the solute
• Solution: the homogeneous mixture of solute and solvent

By using molarity, you can easily compare different solutions and understand their relative concentration. This is critical when performing chemical reactions or preparing laboratory solutions, ensuring desired reactions occur with expected results.

Mole Concept

The mole is a fundamental concept in chemistry, acting as a bridge between the microscopic world of atoms and molecules and the macroscopic world we interact with. A mole is a unit that measures the amount of substance. One mole corresponds to Avogadro’s number, which is approximately \(6.022 \times 10^{23}\) entities, be they atoms, molecules, or particles.

In calculations, the mole concept is pivotal. Molarity uses the mole as its base measurement. For instance, if you want to find the molarity, you need to know how many moles of a substance are present in a particular volume of solution.

Here’s the simple formula for calculating moles:

• Mass of the substance divided by its molar mass: \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \)

Mastering the mole concept is crucial for any chemistry student, as it provides the quantitative foundation necessary for understanding chemical reactions.

Volume Conversion

In chemistry, accurately converting volume units is often necessary to ensure correct solution concentration calculations. Volume can be expressed in different units, but as seen in molarity calculations, volumes are typically needed in liters (L).

A commonly encountered unit is milliliters (mL). Converting between these two is straightforward using the conversion factor:

• 1 milliliter = 0.001 liters
• To convert mL to L, divide by 1000
• Conversely, to convert L to mL, multiply by 1000

This conversion ensures the volume aligns with the molarity formula, making the calculation seamless and reducing errors. Once familiar with these conversions, adjusting the parameters of your solutions becomes second nature, aiding in precise and exact scientific analyses.