Understanding the Problem

We are given 20 mL of pure acetic acid with density 0.75 g/mL, mixed with 50 g of water (density 1 g/mL). We need to find the molality of acetic acid in the final solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent. Here, acetic acid is the solute and water is the solvent.

Step-by-Step Solution

Step 1: Calculate the mass of acetic acid (solute)

Given volume of acetic acid = 20 mL, density = 0.75 g/mL.

Mass of acetic acid = Volume × Density = 20 mL × 0.75 g/mL = 15 g.

Step 2: Identify the mass of solvent (water)

Given mass of water = 50 g. Since molality requires solvent mass in kg, convert grams to kilograms.

Mass of water in kg = 50 g / 1000 = 0.05 kg.

Step 3: Calculate the number of moles of acetic acid

Molar mass of acetic acid (CH₃COOH) = (12×2) + (1×4) + (16×2) = 60 g/mol.

Moles of acetic acid = Mass / Molar mass = 15 g / 60 g/mol = 0.25 mol.

Step 4: Compute the molality

Molality (m) = Moles of solute / Mass of solvent (in kg) = 0.25 mol / 0.05 kg = 5 mol/kg.

Final Answer: The molality of acetic acid in the final solution is $5mol/kg$.

Related Topics

• Concentration Units: Molality, Molarity, Mole Fraction
• Density and Mass-Volume Relationships
• Solution Preparation and Properties

Important Formulae and Theory

Molality (m): $m=\frac{numberofmolesofsolute}{massofsolvent\left(inkg\right)}$

Density: $\rho =\frac{m}{V}$, where ρ is density, m is mass, V is volume.

Moles: $n=\frac{m}{M}$, where n is number of moles, m is mass, M is molar mass.

Note: Molality is temperature-independent because it involves mass, not volume (unlike molarity).