Problem 57 - Chemistry: An Atoms-Focused Approach

Question

Calculate the molality of each of the following solutions: a. 0.875 mol of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in \(1.5 \mathrm{kg}\) of water b. 11.5 mmol of acctic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) in \(65 \mathrm{g}\) of water c. 0.325 mol of baking soda \(\left(\mathrm{NaHCO}_{3}\right)\) in \(290.0 \mathrm{g}\) of water

Step-by-Step Solution

Answer

Question: Determine the molality of each of the following solutions: a) a glucose solution containing 0.875 mol of glucose and 1.5 kg of water; b) an acetic acid solution containing 11.5 mmol of acetic acid dissolved in 65 g of water; c) a baking soda solution with 0.325 mol baking soda dissolved in 290.0 g of water. Answer: a) The molality of the glucose solution is 0.583 mol/kg. b) The molality of the acetic acid solution is 0.177 mol/kg. c) The molality of the baking soda solution is 1.12 mol/kg.

Step 1: Solution a. Molality of glucose solution

First, find the moles of solute which is given as 0.875 mol. Then, find the mass of solvent, which is given as 1.5 kg. Next, use the formula to calculate molality: molality = (0.875 mol) / (1.5 kg) = 0.583 mol/kg. So, the molality of the glucose solution is 0.583 mol/kg.

Step 2: Solution b. Molality of acetic acid solution

First, convert the 11.5 mmol of acetic acid to mol by dividing by 1000: 11.5 mmol / 1000 = 0.0115 mol. Then, find the mass of solvent by converting the 65 g of water to kg: 65 g / 1000 = 0.065 kg. Next, use the formula to calculate molality: molality = (0.0115 mol) / (0.065 kg) = 0.177 mol/kg. So, the molality of the acetic acid solution is 0.177 mol/kg.

Step 3: Solution c. Molality of baking soda solution

First, find the moles of solute which is given as 0.325 mol. Then, find the mass of solvent, which is given as 290.0 g. Convert the mass to kg: 290 g / 1000 = 0.290 kg. Next, use the formula to calculate molality: molality = (0.325 mol) / (0.290 kg) = 1.12 mol/kg. So, the molality of the baking soda solution is 1.12 mol/kg.

Key Concepts

Solution Chemistry

Solution chemistry involves studying the properties and behaviors of substances dissolved in a solvent, creating a solution. When calculating the concentration of a solution, molality is a commonly used measure. Molality is expressed in terms of the amount of solute per mass of solvent and is represented by the formula:

• Molality = moles of solute / kilograms of solvent.

This unit is often useful when the temperature changes because molality does not depend on temperature, unlike molarity which depends on the volume of the solution.
In practical applications, molality helps in understanding how different substances react with solvents. The consistent measure it provides, regardless of external conditions, makes it essential for accurate chemical analysis.

Moles and Mass

Understanding the concept of moles and mass is crucial in solution chemistry. A mole is a standard unit in chemistry that quantifies the amount of a substance. It is similar to counting by dozens but on a much larger scale (Avogadro's number, which is approximately \(6.022 \times 10^{23}\)).
Calculating molality requires knowledge of both moles and mass. To convert into moles, you can use the equation:

• Number of moles = mass of substance (g) / molar mass (g/mol).

This conversion is important in calculating molality, as it ensures that the amount of solute in moles is appropriately correlated to the mass of the solvent in kilograms.
Understanding these concepts allows chemists to precisely prepare solutions with desired concentrations.

Unit Conversion

Unit conversion is essential in chemistry, particularly when dealing with measurements ranging in different units such as molality, which uses moles and kilograms.
In the provided problems, conversion between grams and kilograms was necessary. Remember, converting mass from grams to kilograms is achieved by dividing by 1000:

• 1 g = 0.001 kg.

Similarly, when dealing with millimoles (mmol), it is crucial to convert them to moles by dividing by 1000 as well:

• 1 mmol = 0.001 mol.

Such conversions are vital as they ensure calculations are in the required units, avoiding potential errors in your measurements and results. Multiple unit systems can be challenging, but mastering them ensures accuracy in your chemical calculations.