Question

Mass of ethanoic acid required to prepare 0.5m solution containing 100g of water?

Hint:

Always double-check that your solvent mass is converted to kilograms when working with molality equations.
A common mistake is using grams directly, which offsets the final answer by a factor of 1000.

Answer 1

Step 1: Understanding the Concept:
Molality (\( m \)) is a concentration term defined as the number of moles of solute dissolved per kilogram of the solvent.
Step 2: Key Formula or Approach:
The relevant formulas are:
\[ \text{Molality } (m) = \frac{\text{Moles of solute } (n)}{\text{Mass of solvent in kg}} \] \[ \text{Moles of solute } (n) = \frac{\text{Mass } (W)}{\text{Molar Mass } (M)} \] Step 3: Detailed Explanation:
The solvent here is water, and its mass is given as 100 g.
First, convert this mass into kilograms:
\[ \text{Mass of solvent} = \frac{100}{1000} = 0.1 \text{ kg} \] The target molality (\( m \)) of the solution is given as 0.5 m (mol/kg).
Now, calculate the required number of moles (\( n \)) of the solute (ethanoic acid):
\[ 0.5 = \frac{n}{0.1} \] \[ n = 0.5 \times 0.1 = 0.05 \text{ moles} \] Next, calculate the molar mass of ethanoic acid (\( \text{CH}_3\text{COOH} \)).
Using the atomic masses: Carbon (C) = 12, Hydrogen (H) = 1, Oxygen (O) = 16.
\[ M = 12 + (3 \times 1) + 12 + (2 \times 16) + 1 \] \[ M = 12 + 3 + 12 + 32 + 1 = 60 \text{ g/mol} \] Now, calculate the mass (\( W \)) of ethanoic acid corresponding to 0.05 moles:
\[ W = n \times M \] \[ W = 0.05 \text{ mol} \times 60 \text{ g/mol} \] \[ W = 3 \text{ g} \] Step 4: Final Answer:
The mass of ethanoic acid required is 3 g.