What would be the molality of 60% (mass/mass) aqueous solution of `CH_(3)COOH`? (Molar mass of `CH_(3)COOH = 60 g mol^(-1)` )

Explanation

To find the molality of a 60% (mass/mass) aqueous solution of acetic acid (CH₃COOH), we can follow these steps:

### Step 1: Understand the Definition of Molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula is:
\[ \text{Molality} (m) = \frac{\text{Number of moles of solute}}{\text{Mass of solvent (kg)}} \]

### Step 2: Determine the Mass of Solute and Solvent
Given that the solution is 60% acetic acid by mass, we can assume we have 100 g of the solution for simplicity. This means:
- Mass of acetic acid (solute) = 60 g
- Mass of water (solvent) = 100 g - 60 g = 40 g

### Step 3: Convert the Mass of Solvent to Kilograms
Since molality requires the mass of the solvent in kilograms, we convert 40 g to kg:
\[ \text{Mass of solvent} = 40 \text{ g} = 0.040 \text{ kg} \]

### Step 4: Calculate the Number of Moles of Solute
To find the number of moles of acetic acid, we use the formula:
\[ \text{Number of moles} = \frac{\text{Mass of solute (g)}}{\text{Molar mass (g/mol)}} \]
Given that the molar mass of acetic acid (CH₃COOH) is 60 g/mol:
\[ \text{Number of moles} = \frac{60 \text{ g}}{60 \text{ g/mol}} = 1 \text{ mole} \]

### Step 5: Calculate Molality
Now we can substitute the values into the molality formula:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent (kg)}} = \frac{1 \text{ mole}}{0.040 \text{ kg}} = 25 \text{ mol/kg} \]

### Final Answer
The molality of the 60% (mass/mass) aqueous solution of acetic acid is:
\[ \text{Molality} = 25 \text{ mol/kg} \]

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