Question

What volume of solution is needed to provide \(0.229 \mathrm{~g}\) of \(\mathrm{CH}_{2} \mathrm{O}\) from a \(0.00560 \mathrm{M}\) solution?

Short answer

You need approximately 1.36 liters of the solution.

Step-by-step solution

Understanding the Exercise

We need to find the volume of a solution required to provide a specified mass of \(\mathrm{CH}_2 \mathrm{O}\). Given a molarity (M) of \(0.00560\) and mass \(\mathrm{CH}_2 \mathrm{O}\) required is \(0.229\) grams.

Convert Mass to Moles

First, we need to convert the given mass of \(\mathrm{CH}_2 \mathrm{O}\) to moles. We use the molar mass of \(\mathrm{CH}_2 \mathrm{O}\), which is \(12.01 + 2 imes 1.01 + 16.00 = 30.03\) g/mol. The number of moles, \(n\), is calculated as: \(n = \frac{0.229 \, \text{g}}{30.03 \, \text{g/mol}}\).

Calculate Number of Moles

Perform the division to find the number of moles: \(n = \frac{0.229}{30.03} \, \text{moles} \), which approximates to \(0.00763 \, \text{moles}\).

Use Molarity to Find Volume

Using the molarity formula \(\text{M} = \frac{\text{moles}}{\text{liters}}\), rearrange to solve for volume in liters: \(\text{Volume} = \frac{\text{moles}}{\text{M}}\). Substitute in the values: \(\text{Volume} = \frac{0.00763 \, \text{moles}}{0.00560 \, \text{M}}\).

Calculate the Required Volume

Perform the division to calculate the volume needed in liters: \(\text{Volume} = \frac{0.00763}{0.00560} \, \text{liters} \), which equals approximately \(1.36 \, \text{liters}\).

Key concepts

Molarity Calculation

Molarity is a fundamental concept in chemistry that helps us understand the concentration of a solution. It is defined as the number of moles of solute per liter of solution. In simple terms, molarity tells you how much solute is present in a specific volume of liquid.
To calculate molarity, use the formula:

• Molarity (M) = \( \frac{\text{moles of solute}}{\text{liters of solution}} \)

In problems like the one presented in the exercise, understanding molarity allows us to find either the volume of the solution required or to determine the concentration of a solution given the amount of solute.
Once you know the molarity, you can use it to find out other quantities as needed, whether it's adapting for different volume of solutions or concentration changes.

Mass to Moles Conversion

Converting mass to moles is a crucial step in this exercise that involves linking a substance's mass to its chemical representation in moles. To do this, you need to know the molar mass of the substance, which is the mass of one mole of that substance expressed in grams.
The formula used for this conversion is:

• Number of moles (n) = \( \frac{\text{mass of the substance in grams}}{\text{molar mass in grams/mole}} \)

In our problem, for instance, we calculated the moles of \( \text{CH}_2\text{O} \) using its molar mass of 30.03 g/mol. By dividing the given mass, 0.229 grams, by this molar mass, you obtain the moles of \( \text{CH}_2\text{O} \), necessary for further calculations.
This conversion forms a bridge between the tangible mass measurements and the molar units needed for stoichiometric computations.

Volume Determination

After obtaining the number of moles, the next step is to determine the volume of the solution needed. This is where the concept of molarity proves useful once again. By rearranging the definition of molarity, you can solve for volume:

• Volume (L) = \( \frac{\text{moles of solute}}{\text{molarity (M)}} \)

In this calculation, you simply divide the number of moles of \( \text{CH}_2\text{O} \) by the solution's molarity, 0.00560 M, which results in the volume required.
Determining volumes like this is essential in preparing solutions with precise concentrations, especially in laboratory settings where exact measurements are paramount. With practice, these calculations become intuitive, allowing for efficient and accurate solution preparation.
Remember, the ultimate goal is to ensure the solution has the desired characteristics, making volume determination a key aspect of solution stoichiometry.