Problem 14 - Chemistry Matter and Change

Question

What is the percent by volume of isopropyl alcohol in a solution that contains 24 \(\mathrm{mL}\) of isopropyl alcohol in 1.1 \(\mathrm{L}\) of water?

Step-by-Step Solution

Answer

The percent by volume of isopropyl alcohol is approximately 2.14%.

Step 1: Convert the Volume of Water to Milliliters

First, convert the volume of water from liters to milliliters because the percent by volume is typically calculated using the same units. Since 1 liter equals 1,000 milliliters, multiply 1.1 liters by 1,000 to get 1,100 milliliters.

Step 2: Find the Total Volume of the Solution

Add the volume of isopropyl alcohol to the volume of water to find the total volume of the solution. Here, add 24 mL (isopropyl alcohol) to 1,100 mL (water) to get a total volume of 1,124 mL.

Step 3: Calculate the Percent by Volume

Use the formula for percent by volume: \( \text{Percent by volume} = \left( \frac{\text{Volume of solute}}{\text{Total volume of solution}} \right) \times 100 \). Substituting the values: \( \text{Percent by volume} = \left( \frac{24}{1,124} \right) \times 100 \approx 2.14\% \).

Key Concepts

Solution Concentration

Solution concentration is an essential concept in chemistry when dealing with mixtures. It tells us how much of one substance, the solute, is present compared to the entire mixture, known as the solution. When we talk about solution concentration in terms of percent by volume, we're specifically referring to how many milliliters of a solute exists per 100 milliliters of the total solution.

• The formula used is: \[ \text{Percent by volume} = \left( \frac{\text{Volume of solute}}{\text{Total volume of solution}} \right) \times 100 \]
• This allows us to express the concentration as a percentage, a very intuitive way to compare concentrations of various solutions.

Understanding percent by volume is crucial because it provides a straightforward measure of how concentrated a solution is, which is important for both practical and theoretical purposes in chemical mixtures.

Isopropyl Alcohol

Isopropyl alcohol, also known as
d2-propanol, is a widely used solvent in chemical laboratories and everyday applications. It is a clear, colorless liquid with a strong odor, and it evaporates quickly.

• Commonly used in medical and household cleaning products.
• It acts as an effective disinfectant and is used in sanitizers.
• In chemical calculations, it serves as a common example of a solute in various types of solutions.

In our exercise, we are interested in finding out how much isopropyl alcohol contributes to the overall solution by measuring its percent by volume. This measure helps assess the strength or effectiveness of the solution when used as a disinfectant or cleaning agent.

Volume Conversion

Volume conversion is a fundamental skill in chemistry that assists in making accurate calculations. It's essential because solution concentrations usually require consistent units of measurement.
When working with liquids, conversions often involve moving between liters and milliliters.

• 1 liter (L) equals 1,000 milliliters (mL).
• To convert liters to milliliters, multiply by 1,000: \[ 1.1 \text{ L} \times 1,000 = 1,100 \text{ mL} \]
• Having the same unit allows for simple addition or subtraction and straightforward percent calculation.

In our context, converting water volume from liters to milliliters ensures that both the solute and solvent measurements are in milliliters, thereby allowing easy calculation of total solution volume essential for determining percent by volume.

Chemical Calculations

Chemical calculations are a set of mathematical techniques applied to solve problems related to chemical reactions and solutions. They involve using formulas and converting units to find precise concentrations and measurements.

• Understanding chemical calculations allows for precise preparation of recipes in lab settings or industrial applications.
• They ensure proper proportions of substances, which is critical for safety and efficacy.
• For percent by volume, it directly involves the calculation: \[ \text{Percent by Volume} = \left( \frac{24 \text{ mL of isopropyl}}{1,124 \text{ mL total solution}} \right) \times 100 \approx 2.14\% \]

This particular judgment shows how even small volumes of substances can have significant impacts on solution concentrations, underlying the importance of accuracy in chemical measurements.