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 (m) of acetic acid (CH3COOH), we first need to determine the number of moles of acetic acid in 1 liter of the solution.
Given:
Molarity (M) = 2 M = 2 mol/L
Density = 1.20 g/mL
To calculate the number of moles:
1. Calculate the mass of 1 L of the solution:
Mass = Density × Volume
= 1.20 g/mL × 1000 mL
= 1200 g
2. Calculate the number of moles of acetic acid using its molar mass:
Molar mass of CH3COOH = (12.01 g/mol × 2) + (1.01 g/mol × 4) + (16.00 g/mol) + (1.01 g/mol) = 60.05 g/mol
Moles = Mass / Molar mass
= 1200 g / 60.05 g/mol
≈ 19.98 mol
Now that we have the number of moles in 1 liter of solution, we can calculate the molality (m) using the following formula:
m = moles of solute / mass of the solvent (in kg)
3. Convert the mass of solvent from grams to kilograms:
Mass of solvent = Mass - Mass of solute
= 1200 g - 60.05 g ≈ 1139.95 g ≈ 1.14 kg
4. Calculate the molality:
m = moles of solute / mass of the solvent
= 19.98 mol / 1.14 kg
≈ 17.51 mol/kg
Therefore, the molality of the acetic acid solution is approximately 17.51 mol/kg.