Question

Calculate the molarity and mole fraction of acetone in a \(1.00-m\) solution of acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right)\) in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right) .\) (Density of acetone \(=0.788 \mathrm{g} / \mathrm{cm}^{3} ;\) density of ethanol \(=0.789\) \(\mathrm{g} / \mathrm{cm}^{3} .\) ) Assume that the volumes of acetone and ethanol add.

Short answer

The molarity of acetone in the solution is \(0.931\,M\) and the mole fraction of acetone in the mixture is \(0.0552\).

Step-by-step solution

Calculate the mass of each component in the solution

Using the given 1.00-molar solution of acetone and ethanol, we start by determining the mass of each component. Recall that the molarity (M) is defined as moles of solute per liter of solvent. We also use the given density values to convert the mass of each component to volume. 1. Using the given molarity of the solution (\(1.00\,M\)) which means \(1.00\, mol\) of acetone is in \(1\,L\) of ethanol. 2. Look up the molar masses of acetone \((CH_{3}COCH_{3})\) and ethanol \((C_{2}H_{5}OH)\): Molar mass of acetone = \(58.08\,g/mol\) Molar mass of ethanol = \(46.07\,g/mol\) 3. Calculate the total mass of each component in grams using the molar masses: Mass of acetone = moles of acetone × molar mass of acetone Mass of acetone = \(1.00\,mol \times 58.08\,g/mol = 58.08\,g\) 4. Since there is \(1\,L\) of ethanol, so the mass of ethanol can be calculated using density: Mass of ethanol = Density × Volume Mass of ethanol = \(0.789\, g/cm^3 \times 1000\, cm^3 = 789\, g\)

Calculate the moles of acetone and ethanol

Now, we can calculate the moles of acetone and ethanol using the calculated masses and molar masses. 1. Moles of acetone = \(\frac{mass\,of\,acetone}{molar\,mass\,of\,acetone} = \frac{58.08\,g}{58.08\,g/mol} = 1.00\,mol\) 2. Moles of ethanol= \(\frac{mass\,of\,ethanol}{molar\,mass\,of\,ethanol} = \frac{789\,g}{46.07\,g/mol} = 17.12\,mol\)

Calculate the molarity of acetone

Now that we know the moles of acetone and the volume of the solution in liters, we can find the molarity of acetone. Molarity of acetone = \(\frac{moles\,of\,acetone}{volume\,of\,solution\,in\,liter}\) Since the problem states that the volumes of acetone and ethanol add, we need to calculate the total volume of the solution using the mass and density: 1. Volume of acetone = \(\frac{mass\,of\,acetone}{density\,of\,acetone} = \frac{58.08\,g}{0.788\,g/cm^3} = 73.66\,cm^3\) 2. Volume of ethanol = \(\frac{mass\,of\,ethanol}{density\,of\,ethanol} = \frac{789\,g}{0.789\,g/cm^3} = 1000\,cm^3\) 3. Total volume of solution = volume of acetone + volume of ethanol = \(73.66\,cm^3 + 1000\,cm^3 = 1073.66\,cm^3 = 1.07366\,L\) Molarity of acetone = \(\frac{1.00\,mol}{1.07366\,L} = 0.931\,M\)

Calculate the mole fraction of acetone

Now that we know the moles of both components, we can calculate the mole fraction of acetone by using the following formula: Mole fraction of acetone = \(\frac{moles\,of\,acetone}{total\,moles\,of\,mixture} = \frac{1.00\,mol}{1.00\,mol + 17.12\,mol}\) Mole fraction of acetone = \(\frac{1.00\,mol}{18.12\,mol} = 0.0552\) So, the molarity of acetone in the solution is \(0.931\,M\) and the mole fraction of acetone in the mixture is \(0.0552\).

Key concepts

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is the ratio of the number of moles of a particular substance to the total number of moles of all the substances present in the mixture. Since mole fractions are ratios, they are unitless and indicate the proportion of one component in the mixture.

The mole fraction is especially useful because it is independent of temperature and pressure. In our context, the mole fraction of acetone in an acetone-ethanol mixture helps us understand the relative amount of acetone compared to the total mixture. For instance, a mole fraction of 0.0552 for acetone suggests that acetone makes up 5.52% of the total number of moles in the mixture.

Acetone in Ethanol Solution

When acetone (CH_3COCH_3) is mixed with ethanol (C_2H_5OH), it forms a solution where acetone is the solute, and ethanol acts as the solvent. The chemical and physical properties of each component affect how they interact with each other.

For instance, the density of each liquid is used to translate between volume and mass—a critical step in calculations involving solutions. It's important to note that even though chemists often assume additivity of volumes when components form a solution, this is an approximation, as the actual volume can sometimes be slightly less or more than the sum of individual volumes. Knowledge of the molar masses of acetone and ethanol is also crucial in calculating the molarity and the mole fraction, as seen in the step-by-step solution.

Solution Concentration

Solution concentration quantifies the amount of solute in a given amount of solvent or solution. There are various ways to express concentration, with molarity being one of the most common. Molarity (M) is defined as moles of solute per liter of solution, and it is a standard unit of concentration in chemistry for solution reactions.

In our exercise, the molarity of acetone in the ethanol solution provides information on how strong or weak the solution is. A molarity of 0.931 M signifies that there are 0.931 moles of acetone per liter of the ethanol solution. This is a crucial measure when reacting this solution with other substances, as it determines how much reactant is needed and the expected yield of the products. To ensure a complete understanding, students should practice converting between mass, moles, volume, and molarity using known densities and molar masses.