Question

What are the molarities of the following solutes? (a) aspartic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{5} \mathrm{NO}_{4}\right)\) if \(0.405 \mathrm{g}\) is dissolved in enough water to make \(100.0 \mathrm{mL}\) of solution (b) acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O},(d=0.790 \mathrm{g} / \mathrm{mL})\) if \(35.0 \mathrm{mL}\) is dissolved in enough water to make \(425 \mathrm{mL}\) of solution (c) diethyl ether, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O},\) if \(8.8 \mathrm{mg}\) is dissolved in enough water to make 3.00 L of solution

Short answer

The molarities of the solutions are as follows: (a) Aspartic acid, 0.025 M; (b) Acetone, 1.02 M; (c) Diethyl ether, 0.00004 M

Step-by-step solution

Calculate the molarity of aspartic acid

While normally we would convert grams to moles using the molar mass of aspartic acid, we are given the total mass is 0.405g, so we can use that directly. Convert the volume of the solution to liters by dividing it by 1000. Therefore, the volume of the solution is 0.1L. Then use the definition of molarity, which is moles of solute per liter of solution. Given the density of aspartic acid is 1.66 g/cm^3, the molarity can be calculated as follows: \[ M = \frac{0.405 g}{(1.66 g/mol) \times 0.1 L} \]

Calculate the molarity of acetone

Calculate the number of moles of acetone by multiplying its volume by its density and dividing by its molar mass: \[ moles = \frac{35.0 mL \times (0.790 g/mL)}{(58.08 g/mol)} \] The volume of the solution is 425 mL, or 0.425 L. Therefore, the molarity of the acetone solution can be calculated as: \[ M = \frac{moles}{0.425 L} \]

Calculate the molarity of diethyl ether

Calculate the number of moles of diethyl ether by converting the mass of the solute from mg to g, and then dividing by the molar mass: \[ moles = \frac{8.8 mg \times (1 g/1000 mg)} {(74.12 g/mol)} \] The volume of the solution is 3.00 L. Therefore, the molarity of the diethyl ether solution can be calculated as: \[ M = \frac{moles}{3.00 L} \] The molarity for each solution has been determined.

Key concepts

Aspartic Acid Solution

To begin calculating the molarity of an aspartic acid solution, we need to understand the relationship between mass, molarity, and volume. Aspartic acid, with the formula \( \mathrm{H}_2 \mathrm{C}_4 \mathrm{H}_5 \mathrm{NO}_4 \), is dissolved into a solution. We know that molarity \( (M) \) is defined as the number of moles of solute divided by the liters of solution.
We are given that 0.405 grams of aspartic acid is dissolved in water to make a 100.0 mL solution. Our first step is to convert the volume from milliliters to liters by dividing by 1000, so our volume of solution is 0.100 L.
Next, we need the molar mass of aspartic acid, which is approximately 133.10 g/mol for calculations. Now, the equation for molarity \( M \) is:
\[ M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume of solution (L)}} \]
Substitute the values into the formula to find the molarity. With these calculations, you will successfully determine how concentrated your aspartic acid solution is.

Acetone Molarity

Understanding how to calculate the molarity of acetone involves similar steps to those used for aspartic acid.
Acetone, \( \mathrm{C}_3 \mathrm{H}_6 \mathrm{O} \), is often used in laboratories, so it's important to find how concentrated it is in your solution.
Given 35.0 mL of acetone (with density \( 0.790 \mathrm{g/mL} \)) is dissolved, we need to convert this volume to a mass. Using the density, multiply the volume by the density to get the mass.

• Convert 35.0 mL of acetone to grams: \( 35.0 \text{ mL} \times 0.790 \text{ g/mL} = 27.65 \text{ g} \)

Next, we calculate the moles of acetone by dividing the mass by the molar mass, approximately 58.08 g/mol:

• \( \text{moles} = \frac{27.65 \text{ g}}{58.08 \text{ g/mol}} \)

Then, use the definition of molarity, with the volume of the solution converted to liters, 0.425 L in this case.
Hence, molarity \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \) is calculated, giving the concentration of your acetone solution.

Diethyl Ether Molarity

Calculating the molarity of diethyl ether involves understanding a slightly different unit conversion. Diethyl ether, represented chemically as \( (\mathrm{C}_2 \mathrm{H}_5)_2 \mathrm{O} \), is a common solvent.
For this calculation, the initial mass of diethyl ether is given in milligrams, specifically 8.8 mg. Begin by converting this mass into grams by dividing by 1000.

• 8.8 mg converted to grams is \( 8.8 \times 10^{-3} \text{ g} \).

Next, we find the moles of diethyl ether using its molar mass, which is roughly 74.12 g/mol.

• \( \text{moles} = \frac{8.8 \times 10^{-3} \text{ g}}{74.12 \text{ g/mol}} \)

Then we compute the molarity by dividing the number of moles by the volume of the solution in liters, which is 3.00 L in this scenario.
This step-by-step calculation will yield the molarity, giving insight into how dilute or concentrated the diethyl ether solution is.