The density of a 2.05M solution of acetic acid in water 1.02g/mL the molality of solution is

To find the molality of a 2.05 M solution of acetic acid with a density of 1.02 g/mL, we need to follow a few steps that involve understanding the definitions of molarity and molality, as well as some basic calculations. Let’s break it down step by step.

Understanding Molarity and Molality

Molarity (M) is defined as the number of moles of solute per liter of solution. In this case, we have a 2.05 M solution, which means there are 2.05 moles of acetic acid in every liter of solution. On the other hand, molality (m) is defined as the number of moles of solute per kilogram of solvent. To convert from molarity to molality, we need to know the mass of the solvent in kilograms.

Step 1: Calculate the Mass of the Solution

Since we know the density of the solution, we can calculate the mass of 1 liter of this solution. The density is given as 1.02 g/mL, which means:

• 1 liter = 1000 mL
• Mass of 1 liter of solution = Density × Volume = 1.02 g/mL × 1000 mL = 1020 g

Step 2: Calculate the Mass of Acetic Acid

Next, we need to find out how much of that mass is acetic acid. We already know that there are 2.05 moles of acetic acid in 1 liter of solution. The molar mass of acetic acid (CH₃COOH) is approximately 60.05 g/mol. Therefore, the mass of acetic acid in the solution is:

• Mass of acetic acid = Moles × Molar Mass = 2.05 moles × 60.05 g/mol ≈ 123.10 g

Step 3: Calculate the Mass of the Solvent (Water)

Now, we can find the mass of the solvent (water) by subtracting the mass of acetic acid from the total mass of the solution:

• Mass of water = Mass of solution - Mass of acetic acid = 1020 g - 123.10 g ≈ 896.90 g

To convert this mass into kilograms, we divide by 1000:

• Mass of water in kg = 896.90 g / 1000 ≈ 0.897 kg

Step 4: Calculate the Molality

Finally, we can calculate the molality using the formula:

• Molality (m) = Moles of solute / Mass of solvent in kg
• Molality = 2.05 moles / 0.897 kg ≈ 2.28 m

Final Result

The molality of the 2.05 M solution of acetic acid is approximately 2.28 m. This means that in every kilogram of water, there are about 2.28 moles of acetic acid dissolved. Understanding these calculations helps in various applications, such as in chemistry labs or when preparing solutions for experiments.