Problem 59 - Chemistry: An Atoms-Focused Approach

Question

Calculate the molality of each of the following aqueous solutions: a. \(1.30 M \mathrm{CaCl}_{2}(d=1.113 \mathrm{g} / \mathrm{mL})\) b. \(2.02 M\) fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}, d=1.139 \mathrm{g} / \mathrm{mL}\right)\) c. \(8.94 M\) ethylene glycol (antifreeze, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\), \(d=1.069 \mathrm{g} / \mathrm{mL})\) d. \(1.97 M\) LiCl \((d=1.046 \mathrm{g} / \mathrm{mL})\)

Step-by-Step Solution

Answer

In summary, we calculated the molality of each solution as follows: a) 1.34 mol/kg for the \(1.30 M \mathrm{CaCl}_{2}\) solution. b) 2.61 mol/kg for the \(2.02 M\) fructose solution. c) 17.4 mol/kg for the \(8.94 M\) ethylene glycol solution. d) 2.05 mol/kg for the \(1.97 M\) LiCl solution.

Step 1: Moles of Solute CaCl2

Using the molarity, we first calculate the moles of CaCl2 as follows: moles of CaCl2 = molarity × volume, Assume volume = 1 L, moles of CaCl2 = 1.30 mol

Step 2: Mass of Solution

Now we will use the density to find the mass of the solution: mass of solution = volume × density mass of solution = 1000 mL × 1.113 g/mL = 1113 g

Step 3: Mass of Solvent

Next, we need to subtract the mass of CaCl2 from the mass of the solution to find the mass of the solvent: mass of CaCl2 = moles of CaCl2 × molar mass of CaCl2 = 1.30 mol × 111 g/mol = 144.3 g mass of solvent = mass of solution - mass of CaCl2 = 1113 g - 144.3 g = 968.7 g

Step 4: Calculate Molality

Finally, we will calculate the molality of the solution: molality = moles of solute / mass of solvent (in kg) molality = 1.30 mol / 0.9687 kg = 1.34 mol/kg For solution a, the molality is 1.34 mol/kg. ##Solution b: \(2.02 M\) fructose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}, d=1.139 \mathrm{g} / \mathrm{mL}\right)$##

Step 1: Moles of Solute Fructose

Using the molarity, we calculate the moles of fructose: moles of fructose = 2.02 mol (assuming 1 L of solution)

Step 2: Mass of Solution

Now we will use the density to find the mass of the solution: mass of solution = 1000 mL × 1.139 g/mL = 1139 g

Step 3: Mass of Solvent

Next, we need to subtract the mass of fructose from the mass of the solution to find the mass of the solvent: mass of fructose = moles of fructose × molar mass of fructose = 2.02 mol × 180.16 g/mol = 363.92 g mass of solvent = mass of solution - mass of fructose = 1139 g - 363.92 g = 775.08 g

Step 4: Calculate Molality

Finally, we will calculate the molality of the solution: molality = moles of solute / mass of solvent (in kg) molality = 2.02 mol / 0.77508 kg = 2.61 mol/kg For solution b, the molality is 2.61 mol/kg. ##Solution c: \(8.94 M\) ethylene glycol (antifreeze, $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\(, \)d=1.069 \mathrm{g} / \mathrm{mL})$## We will follow the same steps as before to find the molality of the ethylene glycol solution:

Step 1: Moles of Solute Ethylene Glycol

Using the molarity, we calculate the moles of ethylene glycol: moles of ethylene glycol = 8.94 mol (assuming 1 L of solution)

Step 2: Mass of Solution

Now we will use the density to find the mass of the solution: mass of solution = 1000 mL × 1.069 g/mL = 1069 g

Step 3: Mass of Solvent

Next, we need to subtract the mass of ethylene glycol from the mass of the solution to find the mass of the solvent: mass of ethylene glycol = moles of ethylene glycol × molar mass of ethylene glycol = 8.94 mol × 62.07 g/mol = 555.11 g mass of solvent = mass of solution - mass of ethylene glycol = 1069 g - 555.11 g = 513.89 g

Step 4: Calculate Molality

Finally, we will calculate the molality of the solution: molality = moles of solute / mass of solvent (in kg) molality = 8.94 mol / 0.51389 kg = 17.4 mol/kg For solution c, the molality is 17.4 mol/kg. ##Solution d: \(1.97 M\) LiCl \((d=1.046 \mathrm{g} / \mathrm{mL})\)## We will follow the same steps as before to find the molality of the LiCl solution:

Step 1: Moles of Solute LiCl

Using the molarity, we calculate the moles of LiCl: moles of LiCl = 1.97 mol (assuming 1 L of solution)

Step 2: Mass of Solution

Now we will use the density to find the mass of the solution: mass of solution = 1000 mL × 1.046 g/mL = 1046 g

Step 3: Mass of Solvent

Next, we need to subtract the mass of LiCl from the mass of the solution to find the mass of the solvent: mass of LiCl = moles of LiCl × molar mass of LiCl = 1.97 mol × 42.39 g/mol = 83.51 g mass of solvent = mass of solution - mass of LiCl = 1046 g - 83.51 g = 962.49 g

Step 4: Calculate Molality

Finally, we will calculate the molality of the solution: molality = moles of solute / mass of solvent (in kg) molality = 1.97 mol / 0.96249 kg = 2.05 mol/kg For solution d, the molality is 2.05 mol/kg.

Key Concepts

Molarity

Molarity is a fundamental concept in chemistry that describes the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution. The formula to calculate molarity is:

• \( \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \)

Understanding molarity is crucial as it helps in predicting the behavior of solutions in reactions. For example, a 1.30 M solution of calcium chloride means there are 1.30 moles of calcium chloride in every liter of the solution. This measure helps chemists know how much solute is present in a given volume, which is essential for mixing solutions or performing experiments.

Solute

A solute is a substance that is dissolved in a solvent to form a solution. It can be in the form of solids, liquids, or gases. In our exercises, the solutes are substances like calcium chloride (\(\mathrm{CaCl}_{2}\)), fructose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)), ethylene glycol (\(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)), and lithium chloride (\(\mathrm{LiCl}\)).
The solute concentration is typically measured in moles, which can be calculated using the molarity and the volume of the solution.

• The moles of solute are calculated by multiplying the molarity by the volume of the solution in liters.
• This step is essential for determining other solution properties, like molality or mass percentage, and dictates the quantity of chemical reactions.

Solvent

A solvent is the medium in which the solute is dissolved to form a solution. In most chemical solutions, the solvent is water, making the solution aqueous. The solvent's role is crucial because it determines the solution's properties such as boiling point, freezing point, and vapor pressure.

• To calculate the mass of the solvent, the mass of the solute is subtracted from the total mass of the solution.
• The solvent's mass is necessary for calculating molality, a critical component for understanding the concentration in solution not affected by temperature changes.

Understanding solvent behavior is key in various applications, from pharmaceuticals to industrial processes. Recognizing its mass and interaction with solute highlights its importance in determining solution behaviors.

Density

Density is the measure of mass per unit volume of a substance, typically expressed in grams per milliliter (g/mL) for liquids. It is an essential property used to calculate the mass of a solution given its volume.
In exercises involving molality and molarity, density plays a crucial role. It allows the conversion of volume to mass, which is crucial since molality requires knowing the exact mass of the solvent:

• The mass of a solution is found by multiplying its volume by its density.
• This property is critical for converting between molarity (which depends on volume) and molality (which depends on mass).

For example, knowing the density of a solution like 1.113 g/mL for a calcium chloride solution allows us to calculate the mass needed for further chemical analysis or reactions.