Answer :
The final volume of the solution formed by mixing 100.0 cm³ of water and 100.0 cm³ of methanol at 25°C is approximately 193.1 cm³.
To calculate the volume of the solution formed by mixing 100.0 cm³ of water with 100.0 cm³ of methanol at 25°C, we need to consider the partial molar volumes of each component and the effects of mixing.
Step 1: Understanding Partial Molar Volumes
The partial molar volume of a substance in a mixture is the change in volume of the mixture when one mole of the substance is added, while keeping the temperature and pressure constant. In this case:
- Partial molar volume of water, V_A = 17.7 cm³/mol
- Partial molar volume of methanol, V_B = 38.8 cm³/mol
Step 2: Calculating the Amount of Each Component
To understand how these volumes apply, we first convert the volumes of the liquids we want to mix into moles:
- Density of water (H₂O) = 0.997 g/cm³
- Molar mass of water = 18.02 g/mol
Calculating the moles of water in 100.0 cm³:
[tex]n_A = \frac{100.0 \text{ cm}^3 \times 0.997 \text{ g/cm}^3}{18.02 \text{ g/mol}} \approx 5.53 \text{ mol}[/tex]
- Density of methanol (CH₃OH) = 0.786 g/cm³
- Molar mass of methanol = 32.04 g/mol
Calculating the moles of methanol in 100.0 cm³:
[tex]n_B = \frac{100.0 \text{ cm}^3 \times 0.786 \text{ g/cm}^3}{32.04 \text{ g/mol}} \approx 2.45 \text{ mol}[/tex]
Step 3: Volume Calculation
The total volume of the solution when mixing these two substances is not simply the sum of their volumes (200.0 cm³), because the components will interact and the actual volume of the solution is less due to these interactions.
Step 4: Final Volume of the Mixture
From the data provided, the final volume of the mixture when 100.0 cm³ of water is mixed with 100.0 cm³ of methanol is roughly 193.1 cm³. This is accounting for the intermolecular interactions that occur when the liquids are combined.
