Find out the molality of acetic acid ch3cooh whose molarity is 2 and density of solution is 1.20 g/ ml ?
To find the molality of acetic acid (CH3COOH) given its molarity and the density of the solution, you need to follow a two-step process.
Step 1: Find the moles of acetic acid in the given volume.
Step 2: Calculate the molality using the moles of solute and the mass of the solvent.
Let's go through each step in detail:
Step 1: Find the moles of acetic acid.
Moles = Molarity × Volume (in liters)
Given the molarity of CH3COOH is 2, we can use this value in the formula. To convert the density of the solution from grams to liters, we need to divide the given density by 1000.
Volume = Density of solution (in g/mL) / 1000
Volume = 1.20 g/mL / 1000
Volume = 0.0012 L (or 1.2 mL)
Moles of acetic acid = 2 × 0.0012
Moles of acetic acid = 0.0024 moles
Step 2: Calculate the molality using the moles of solute and the mass of the solvent.
Molality = Moles of solute / Mass of solvent (in kg)
To calculate the mass of the solvent, we need to convert the density of the solution to grams and then to kilograms.
Mass of solvent = Density of solution (in g/mL) × Volume (in mL)
Mass of solvent = 1.20 g/mL × 1.2 mL
Mass of solvent = 1.44 g
Finally, let's calculate the molality:
Molality = 0.0024 moles / 1.44 kg
Molality = 0.00167 mol/kg
Therefore, the molality of acetic acid (CH3COOH) in the given solution is approximately 0.00167 mol/kg.