Find out the molality of acetic acid ch3cooh whose molarity is 2 and density of solution is 1.20 g/ ml ?

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To find the molality of a solute in a solution, we need to know the molarity of the solution and the density of the solvent.

Given:
Molarity of acetic acid (CH3COOH) = 2 M (moles per liter)
Density of the solution = 1.20 g/mL

To calculate the molality, we need to convert the given information into the required units.

Step 1: Convert density from g/mL to g/L
Density of the solution = 1.20 g/mL

There are 1000 mL in a liter, so we can convert the density from g/mL to g/L by multiplying by 1000:
1.20 g/mL x 1000 mL/L = 1200 g/L

Step 2: Calculate the molality using the formula:
Molality (m) = Moles of solute / Mass of solvent (in kg)

We can calculate the moles of acetic acid (CH3COOH) using its molarity and volume of the solution. However, we need to convert the volume from liters to kilograms, considering the density.

To do this, we need to determine the mass of the solution using the following formula:
Mass of the solution (in g) = Volume of the solution (in L) x Density of the solution (in g/L)

Let's assume we have 1 liter (L) of solution:

Mass of the solution = 1 L x 1200 g/L = 1200 g

Now, we can calculate the mass of the solvent (water) using the mass of the solution and mass of the solute:
Mass of the solvent = Mass of the solution - Mass of solute

Since we have the density information, we can calculate the mass of the solute from the molarity:
Mass of the solute (CH3COOH) = Volume of the solution (in L) x Molarity of the solution (in moles/L)

Again, assuming we have 1 liter (L) of solution:

Mass of the solute = 1 L x 2 moles/L = 2 moles

Now, we can calculate the mass of the solvent:
Mass of the solvent = 1200 g - 2 moles = 1198 g

Finally, we can determine the molality:
Molality (m) = Moles of solute / Mass of solvent (in kg)

First, we need to convert the mass of the solvent from grams (g) to kilograms (kg):
Mass of solvent = 1198 g ÷ 1000 = 1.198 kg

Now, we can calculate the molality:
Molality (m) = 2 moles / 1.198 kg ≈ 1.67 m

Therefore, the molality of acetic acid (CH3COOH) in the given solution is approximately 1.67 m.