Determining Solute Amount for Desired Concentration

Understanding Concentration and Dilution Calculations

In chemistry, determining the amount of solute required to achieve a desired concentration in a solution is a fundamental skill. This process involves understanding the relationship between the volume of the solution, the concentration of the solute, and the amount of solute needed. Below are three practical examples that illustrate how to calculate the amount of solute required for a specific concentration.

Example 1: Preparing a Salt Solution for a Laboratory Experiment

In a laboratory setting, a chemist needs to prepare a saline solution with a concentration of 0.9% (w/v) for a cell biology experiment. The chemist wants to make 500 mL of this solution.

To determine the amount of salt (sodium chloride, NaCl) needed:

• Concentration desired: 0.9% (which is 0.9 g of NaCl per 100 mL of solution)
• Volume of solution: 500 mL

The calculation is as follows:

• First, calculate the amount of NaCl needed for 500 mL:

  0.9 g/100 mL × 500 mL = 4.5 g

• Therefore, the chemist needs to dissolve 4.5 grams of NaCl in enough water to make a total volume of 500 mL.

Notes:

• It’s important to ensure that the NaCl is fully dissolved before adding water to reach the final volume.
• If a different volume or concentration is needed, you can adjust the calculation accordingly.

Example 2: Creating a Sugar Solution for a Beverage

A restaurant owner wants to create a sweetened beverage by preparing a sugar solution with a concentration of 10% (w/v). The owner plans to make 1 liter (1000 mL) of this solution.

To find out how much sugar (sucrose, C12H22O11) is required:

• Concentration desired: 10% (meaning 10 g of sugar per 100 mL)
• Volume of solution: 1000 mL

The calculation proceeds as follows:

• First, calculate the total amount of sugar needed:

  10 g/100 mL × 1000 mL = 100 g

• Thus, the restaurant owner will need to dissolve 100 grams of sugar in enough water to total 1 liter.

Notes:

• For better dissolution, it may help to use warm water initially and then cool it down.
• Adjust the concentration calculation if a different sweetness level is desired.

Example 3: Diluting a Concentrated Acid Solution

In an educational setting, a chemistry teacher is preparing a diluted hydrochloric acid (HCl) solution for a demonstration. The teacher has a concentrated solution of 12 M HCl and needs to prepare 250 mL of a 1 M solution.

To determine the amount of concentrated solution needed:

• Concentration of concentrated solution: 12 M
• Desired concentration: 1 M
• Desired volume: 250 mL

Using the dilution formula (C1V1 = C2V2):

• Where C1 = initial concentration, V1 = volume of concentrated solution needed, C2 = final concentration, V2 = final volume of diluted solution.

• Plugging in the values:

  12 M × V1 = 1 M × 250 mL

Solving for V1 gives:

• V1 = (1 M × 250 mL) / 12 M

• V1 = 20.83 mL

• Therefore, the teacher needs to measure out 20.83 mL of the concentrated HCl and dilute it with water to make a final volume of 250 mL.

Notes:

• Always add acid to water, never the reverse, to ensure safety.
• It’s critical to use proper protective equipment when handling concentrated acids.