Question

How many milliliters of a \(12 \%(\mathrm{v} / \mathrm{v})\) propyl alcohol solution would you need to obtain \(4.5 \mathrm{~mL}\) of propyl alcohol?

Short answer

37.5 mL

Step-by-step solution

- Set Up the Problem

Start by identifying the variables given in the problem. You have a 12% (v/v) solution of propyl alcohol, and you need to find how much of this solution you need to get 4.5 mL of pure propyl alcohol.

- Understand the Percentage

The 12% (v/v) means that there are 12 mL of propyl alcohol in every 100 mL of solution. This can be written as a ratio: \( \frac{12 \text{ mL}}{100 \text{ mL}} \text{ solution} \).

- Set Up the Proportion

Let's set up the proportion to solve for the unknown volume of solution (let's call it \( V_s \)). We need to find a volume \( V_s \) such that 12% of it equals 4.5 mL of propyl alcohol. This can be written as: \( \frac{12}{100} \times V_s = 4.5 \).

- Solve for the Volume of the Solution

Rearrange the equation to solve for \( V_s \): \( V_s = \frac{4.5}{\frac{12}{100}} \). Simplify this to: \( V_s = \frac{4.5 \times 100}{12} = 37.5 \text{ mL} \).

Key concepts

volume/volume percentage

Volume/volume percentage (v/v%) is a way to express the concentration of a solution. It tells you how many milliliters of a solute are present in 100 milliliters of the total solution. For example, a 12% (v/v) solution means there are 12 mL of solute (propyl alcohol in this case) in every 100 mL of the solution. To calculate this, you use the formula:

\( \% \text{(v/v)} = \frac{\text{Volume of solute}}{\text{Total volume of solution}} \times 100 \)

This percentage helps in understanding how concentrated the solution is and is particularly useful in chemistry labs for preparing solutions accurately. In this problem, understanding that 12% (v/v) means 12 mL of propyl alcohol in every 100 mL of solution is key. This ratio allows you to set up equations to find the required volumes for other concentrations.

proportions in chemistry

Proportions in chemistry are often used to relate the amounts of different substances in a solution. Proportions can help in solving problems that involve the dilution or mixing of solutions.
When you set up a proportion, you are saying that two ratios are equal. For this problem, the proportion is set up to find the unknown volume of solution that contains a specific amount of solute. The key equation here is:
\( \frac{12}{100} \times V_s = 4.5 \)

Here, 12/100 represents the 12% (v/v) concentration of propyl alcohol. \( V_s \) is the total volume of the solution we need to find, and 4.5 mL is the amount of propyl alcohol needed. By setting up this proportion, you create a relationship that can be solved algebraically to find the unknown value. Proportions are foundational in chemistry because they allow for the calculation of how much of each substance you need when crafting a particular solution.

solution preparation

Preparing a solution with a specific concentration requires careful measurement and mixing of components. It’s essential to understand both the concept of concentration and how to use mathematical tools to achieve the desired solution.
First, define the required concentration and the amount of the solute. For instance, 4.5 mL of propyl alcohol is needed, and the available solution has a concentration of 12% (v/v).
Next, set up an equation or proportion to find the volume of the solution that will contain the necessary amount of solute. Using the steps from the given solution: \( \frac{12}{100} \times V_s = 4.5 \)
Solve for the volume of the solution, \( V_s \), by isolating it on one side of the equation:
\[ V_s = \frac{4.5}{\frac{12}{100}} = \frac{4.5 \times 100}{12} = 37.5 \text{ mL} \]
Therefore, to prepare the solution, measure 37.5 mL of the 12% (v/v) propyl alcohol solution. This precise method of preparation is critical in many scientific and industrial applications to ensure consistent and reliable results.