Question

You need a \(0.050 \mathrm{~mol} / \mathrm{kg}\) aqueous solution of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\). Calculate the mass of methanol you would need to dissolve in \(500 . \mathrm{g}\) water to make this solution.

Short answer

Dissolve 0.801 g of methanol in 500 g of water.

Step-by-step solution

Understand the Mole Concept

Since we need a 0.050 molal solution, it means there are 0.050 moles of solute (methanol) per kilogram of solvent (water). Our solvent is 500 g of water. To use the molal concentration, we first convert the mass of water to kilograms. Since 500 g = 0.5 kg, we can use it in our molality calculations.

Calculate Moles of Methanol Required

Using the molality formula: \[ b = \frac{n}{m} \], where \( b \) is the molality and \( n \) is the number of moles of solute per \( m \) kilograms of solvent, we can rearrange to find \( n \):\[ n = b \times m = 0.050 \times 0.5 = 0.025 \text{ moles} \]. This is the number of moles of methanol needed.

Convert Moles to Mass

To find the mass of methanol, we use the molecular weight of methanol, \( CH_3OH \), which is approximately 32.04 g/mol. We use the formula: \[ ext{mass} = ext{moles} \times ext{molar mass} = 0.025 \times 32.04 = 0.801 ext{ g} \].

Conclusion

You would need to dissolve 0.801 g of methanol in 500 g of water to make a 0.050 molal solution.

Key concepts

Mole Concept

The mole concept is central to understanding chemical reactions and solutions. A "mole" is a unit in chemistry that represents a specific number of particles, typically atoms, molecules, or ions. One mole is equal to Avogadro's number, approximately \(6.022 \times 10^{23}\) particles. This allows chemists to count entities at the molecular level using macroscopic quantities.
When we say a solution has a certain molality, it refers to the number of moles of solute (in this case, methanol) per kilogram of solvent (water). Molality is a measure of concentration that remains constant regardless of temperature changes since it depends only on mass, not volume.
For example, a 0.050 molal solution specifies that there are 0.050 moles of methanol in each kilogram of water. So, to compute the required moles of methanol in a given mass of water, you convert the mass of water from grams to kilograms and apply the molality definition.

Molecular Weight

Understanding molecular weight is fundamental for converting between moles and grams. The molecular weight of a substance is the mass of one mole of its molecules. It is typically expressed in grams per mole (g/mol).
To calculate the molecular weight of methanol (\( \mathrm{CH}_3 \mathrm{OH} \)), you sum the atomic masses of all the atoms in the molecule:

• Carbon (C): 12.01 g/mol
• Hydrogen (H): 1.01 g/mol (for each of the 4 H atoms: 3 in \( \mathrm{CH}_3 \) and 1 in \( \mathrm{OH} \))
• Oxygen (O): 16.00 g/mol

Total molecular weight = \( 12.01 + (4 \times 1.01) + 16.00 = 32.04 \) g/mol.
This number is then used to convert moles to mass. For methanol, having calculated the required number of moles from the given molality and mass of solvent, we can find the necessary grams of methanol using this molecular weight.

Solution Preparation

Preparing a solution with precise concentration requires careful calculations and measurements. It involves several steps, from determining the moles needed to weighing the solute accurately.
Considering the methanol solution, start by identifying the desired molality and the mass of the solvent. We calculate from the given data that we need 0.025 moles of methanol to achieve a 0.050 molal solution with 0.5 kg of water.
Using the molecular weight of methanol, which we established as 32.04 g/mol, we convert the moles to grams. Multiply the number of moles (0.025) by the molecular weight (32.04 g/mol) to find out you need 0.801 grams of methanol.
Finally, dissolve the accurate amount of methanol in the measured mass of water. Make sure to mix the solution well to ensure even distribution of the solute. This precise method guarantees that the resulting solution has the exact concentration required for experiments or reactions.