Question

How many milliliters of \(18.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are required to prepare \(35.0 \mathrm{mL}\) of \(0.250 \mathrm{M}\) solution?

Short answer

Approximately \(0.486 \mathrm{mL}\) of the \(18.0 \mathrm{M}\) \(\mathrm{H}_{2}\mathrm{SO}_{4}\) solution is required to prepare \(35.0 \mathrm{mL}\) of a \(0.250 \mathrm{M}\) solution.

Step-by-step solution

Write down the given values

We have the following information: \(M_1 = 18.0 \mathrm{M}\) (molarity of the concentrated solution) \(M_2 = 0.250 \mathrm{M}\) (molarity of the diluted solution) \(V_2 = 35.0 \mathrm{mL}\) (volume of the diluted solution) We want to find the volume of the concentrated solution, \(V_1\).

Use the dilution formula

The formula that relates the molarity and volume of the concentrated and diluted solution is: \[M_1V_1 = M_2V_2\] Plugging in the known values, we have: \(18.0 \mathrm{M} \times V_1 = 0.250 \mathrm{M} \times 35.0 \mathrm{mL}\)

Solve for V1

To find \(V_1\), we need to isolate \(V_1\) on one side of the equation: \(V_1 = \frac{0.250 \mathrm{M} \times 35.0 \mathrm{mL}}{18.0 \mathrm{M}}\) Now, let's calculate the value of \(V_1\): \(V_1 = \frac{8.75 \mathrm{mL}}{18.0}\)

Calculate the volume of the concentrated solution

Let's calculate the final value of \(V_1\): \(V_1 = 0.486 \mathrm{mL}\) So, roughly \(0.486 \mathrm{mL}\) of the \(18.0 \mathrm{M}\) \(\mathrm{H}_{2}\mathrm{SO}_{4}\) solution is required to prepare \(35.0 \mathrm{mL}\) of a \(0.250 \mathrm{M}\) solution.

Key concepts

Molarity

Understanding the concept of molarity is crucial in the field of chemistry, especially when dealing with solutions. Molarity is defined as the number of moles of solute dissolved per liter of solution, which is usually expressed in units of moles per liter (M). It is given by the formula:

\( Molarity (M) = \frac{moles \, of \, solute}{volume \, of \, solution \, in \, liters} \)

To solve chemistry problems involving molarity, it's important to grasp the relationship between the moles of a substance and its molecular weight, and how to convert between milliliters and liters since molarity depends on volume in liters. In the example given, molarity plays a central role in calculating the volume of concentrated sulfuric acid needed to make a diluted solution.

Solution Preparation

Preparing a solution with a desired molarity involves accurate calculations and precise measurement. When preparing dilute solutions from a more concentrated stock solution, which is often the case in a laboratory, the dilution formula is used to determine the volume of stock solution needed. The key steps include:

• Identifying the required final volume and molarity of the solution.
• Calculating the number of moles of solute needed using the final volume and molarity.
• Determining the volume of the concentrated solution to provide these moles based on its molarity.

With these steps, you can prepare a range of solutions accurately for various experiments and analyses in chemistry.

Chemistry Problem Solving

Tackling chemistry problems like the one presented requires a systematic approach. Efficient problem-solving in chemistry often involves:

• Reading the problem carefully and identifying the knowns and unknowns.
• Choosing the appropriate formula - in this case, the dilution formula.
• Correctly substituting the values into the formula.
• Solving for the unknown using algebraic manipulation if necessary.
• Checking to ensure the solution makes sense both mathematically and conceptually.

By following these steps, students can develop strong analytical skills which are useful for a wide variety of problems in the field of chemistry.

Concentration Dilution

Diluting a concentrated solution to a desired concentration involves the concept of molarity and volume. The dilution formula, which is integral to understanding concentration dilution, is given as:

\( M_1V_1 = M_2V_2 \)

Where:

• \(M_1\): Initial molarity of the concentrated solution
• \(V_1\): Volume of the concentrated solution needed
• \(M_2\): Final molarity of the diluted solution
• \(V_2\): Final volume of the diluted solution

Understanding this principle is fundamental to performing accurate dilutions in chemistry. It ensures that the number of moles of solute remains constant before and after dilution, which is essential in processes like standardizing solutions for titrations or preparing reagents for experimental use.