Problem 100 - General Chemistry: Principles and Modern Applications

Question

When water and methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l}),\) are mixed, the total volume of the resulting solution is not equal to the sum of the pure liquid volumes. (Refer to Exercise 99 for an explanation.) When \(72.061 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and \(192.25 \mathrm{g}\) \(\mathrm{CH}_{3} \mathrm{OH}\) are mixed at \(25^{\circ} \mathrm{C},\) the resulting solution has a density of \(0.86070 \mathrm{g} / \mathrm{mL} .\) At \(25^{\circ} \mathrm{C},\) the densities of water and methanol are \(0.99705 \mathrm{g} / \mathrm{mL}\) and \(0.78706\) \(\mathrm{g} / \mathrm{mL},\) respectively. (a) Calculate the volumes of the pure liquid samples and the solution, and show that the pure liquid volumes are not additive. [ Hint: Although the volumes are not additive, the masses are.] (b) Calculate the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution.

Step-by-Step Solution

Answer

The volume of the methanol-water solution is not equal to the sum of the volumes of pure methanol and water, demonstrating that volumes are not additive in this case. However, the methanol concentration is measured in molarity and is calculated from the number of moles of methanol and the volume of the solution.

Step 1: Calculate the volume of component liquids

Using the formula \( Volume = \frac{Mass}{Density} \), calculate volume of water (\(V_{H2O}\)) as \( \frac{72.061 g}{0.99705 g/mL} \) and volume of methanol (\(V_{CH3OH}\)) as \( \frac{192.25 g}{0.78706 g/mL} \).

Step 2: Calculate total volume of the solution

The total mass of the solution is the sum of the masses of water and methanol i.e. \( 72.061 g + 192.25 g \). Now, using this total mass and the density of the solution, calculate the total volume of the mixture (\(V_{mixture}\)). Use the formula \( Volume = \frac{Mass}{Density} \), so \(V_{mixture}\) = \( \frac{72.061 g + 192.25 g}{0.86070 g/mL} \).

Step 3: Compare the volumes

Now, compare \(V_{H2O} + V_{CH3OH}\) and \(V_{mixture}\). Show that these two volumes are not equal which means the volumes of pure liquids are not additive in mixture.

Step 4: Calculate moles of CH3OH

To calculate the molarity, first, calculate the number of moles of methanol using its given mass and molar mass. The molar mass of \(CH_3OH\) is \(32.04 g/mol\). So, moles of \(CH_3OH\) = \( \frac{192.25 g}{32.04 g/mol} \).

Step 5: Calculate molarity of CH3OH

The molarity of \(CH_3OH\) is defined as the number of moles of \(CH_3OH\) per liter of solution. First convert the volume of the solution (from step 2) from mL to L by dividing by 1000. Now, calculate the molarity as \( \frac{moles of CH_3OH}{Volume of solution in L} \).

Key Concepts

Volume Calculation

To calculate volume, start by using the formula: \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \). This formula allows you to determine the space an object or substance occupies based on its mass and density. In this scenario, the volumes of both water and methanol are determined separately:

• Volume of Water \( (V_{H2O}) \): Given a mass of \( 72.061 \text{ g} \) and a density of \( 0.99705 \text{ g/mL} \), calculate the volume using the formula.
• Volume of Methanol \( (V_{CH3OH}) \): For methanol, use its mass of \( 192.25 \text{ g} \) and density \( 0.78706 \text{ g/mL} \).

By calculating these, you can compare the pure liquid volumes with the final solution volume to determine any contribution to non-additive volume changes.

Non-Additive Volumes

In the context of mixing liquids like water and methanol, you will observe that the volumes do not simply add up. This happens due to the intermolecular interactions which allow the molecules to pack more closely together. When calculating total volume for our solution, we take the combined mass of water and methanol and divide by the solution density:

• Combined Mass: The mass should be the addition of individual masses (\(72.061 \text{ g of } H_2O\) and \(192.25 \text{ g of } CH_3OH\)).
• Solution Volume: Divide this total mass by the density of solution (\(0.86070 \text{ g/mL} \)).

Upon comparison, the sum of the calculated volumes of pure liquids is usually greater than the volume of the mixture, illustrating non-additive volume behavior.

Molarity Calculation

Molarity is a measure that enables you to express the concentration of a solute in a solution. It represents the number of moles of a solute present in a liter of solution. To calculate molarity, you should:

• Calculate Moles of Methanol: Using the molar mass of methanol (\(32.04 \text{ g/mol}\)), convert its mass (\(192.25 \text{ g}\)) into moles: \[ \text{Moles} = \frac{192.25 \text{ g}}{32.04 \text{ g/mol}} \]
• Convert Solution Volume to Liters: Remember to convert from milliliters to liters after finding the solution volume, from the previous non-additive volume step, simply by dividing by 1000.
• Determine Molarity: Use the formula for molarity:\[ \text{Molarity} = \frac{\text{Moles of } CH_3OH}{\text{Volume of Solution in Liters}} \]

This provides you with the concentration of methanol in the solution.

Molar Mass

Molar mass corresponds to the mass of one mole of a substance. For methanol, it's essential to know because it aids in converting the given mass to moles, a crucial step in calculating molarity. Methanol's molecular formula is \(CH_3OH\). Calculating its molar mass involves adding the atomic masses of each atom:

• Carbon \( (C) \): 12.01 g/mol
• Hydrogen \( (H) \): 4 atoms \( \times 1.01 \text{ g/mol} = 4.04 \text{ g/mol} \)
• Oxygen \( (O) \): 16.00 g/mol

Sum these to find the total: \[ \text{Molar Mass of } CH_3OH = 12.01 + 4.04 + 16.00 = 32.05 \text{ g/mol} \]Having the molar mass allows you to complete essential conversions that contribute to accurate concentration calculations.