Problem 46 - Chemistry The Molecular Science

Question

Calculate the molarity of the solute in a solution containing (a) \(14.2 \mathrm{~g} \mathrm{KCl}\) in \(250 . \mathrm{mL}\) solution. (b) \(5.08 \mathrm{~g} \mathrm{~K}_{2} \mathrm{CrO}_{4}\) in \(150 . \mathrm{mL}\) solution. (c) \(0.799 \mathrm{~g} \mathrm{KMnO}_{4}\) in \(400 . \mathrm{mL}\) solution. (d) \(15.0 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in \(500 . \mathrm{mL}\) solution.

Step-by-Step Solution

Answer

(a) 0.760 M, (b) 0.174 M, (c) 0.0126 M, (d) 0.167 M.

Step 1: Determine Molar Mass of Solute

For each solute, calculate the molar mass using the atomic masses of individual elements from the periodic table. - **KCl:** \[ M = 39.1 + 35.5 = 74.6 \text{ g/mol} \]- **\(\text{K}_2\text{CrO}_4\):** \[ M = 2(39.1) + 51.996 + 4(16) = 194.2 \text{ g/mol} \]- **KMnO₄:** \[ M = 39.1 + 54.938 + 4(16) = 158.1 \text{ g/mol} \]- **C₆H₁₂O₆:** \[ M = 6(12) + 12(1) + 6(16) = 180 \text{ g/mol} \]

Step 2: Convert Grams to Moles

Use the formula \(\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\) for each solute:- **KCl:** \[ \text{Moles of KCl} = \frac{14.2}{74.6} = 0.190 \text{ moles} \]- **\(\text{K}_2\text{CrO}_4\):** \[ \text{Moles of \(\text{K}_2\text{CrO}_4\)} = \frac{5.08}{194.2} = 0.0261 \text{ moles} \]- **KMnO₄:** \[ \text{Moles of KMnO₄} = \frac{0.799}{158.1} = 0.00505 \text{ moles} \]- **C₆H₁₂O₆:** \[ \text{Moles of C₆H₁₂O₆} = \frac{15.0}{180} = 0.0833 \text{ moles} \]

Step 3: Convert Solution Volume to Liters

Convert the volume of the solution from milliliters to liters by using the conversion 1 L = 1000 mL.- \(250 \text{ mL} = 0.250 \text{ L}\)- \(150 \text{ mL} = 0.150 \text{ L}\)- \(400 \text{ mL} = 0.400 \text{ L}\)- \(500 \text{ mL} = 0.500 \text{ L}\)

Step 4: Calculate Molarity

Use the formula \(\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}}\):- **KCl:** \[ M = \frac{0.190}{0.250} = 0.760 \text{ M} \]- **\(\text{K}_2\text{CrO}_4\):** \[ M = \frac{0.0261}{0.150} = 0.174 \text{ M} \]- **KMnO₄:** \[ M = \frac{0.00505}{0.400} = 0.0126 \text{ M} \]- **C₆H₁₂O₆:** \[ M = \frac{0.0833}{0.500} = 0.167 \text{ M} \]

Key Concepts

Molar Mass

To calculate molarity, the first step is determining the molar mass of the solute. Molar mass is the weight of one mole of a substance, expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule. For example, potassium chloride (KCl) is composed of one potassium atom and one chloride atom. Using the atomic masses from the periodic table:

• Potassium (K): 39.1 g/mol
• Chlorine (Cl): 35.5 g/mol

Thus, the molar mass of KCl is 39.1 + 35.5 = 74.6 g/mol.
For compounds like potassium chromate ( KₚCrO₄) and glucose (C₆H₁₂O₆), the process is repeated for each atom in the respective chemical formula. Calculating the molar mass is essential because it converts mass measurements into moles, a fundamental unit for chemical reactions.

Solution Preparation

Preparing a solution involves dissolving a known mass of solute into a specific volume of solvent, most often water, to achieve a desired concentration. This requires accurately measuring both the solute and the volume of the solution. The solute is weighed, often in grams, and then transferred into a container where a specific volume of solvent is added.
For example, to prepare a solution of glucose, you might measure 15.0 g of glucose and dissolve it in enough water to make a 500 mL solution. This ensures complete dissolution of the solute to achieve the correct concentration. It's important to account for the total volume of the solution, rather than just the solvent, to ensure precise concentration calculations.
Always remember to convert the solution volume from milliliters to liters, as molarity calculations depend on liters, so 500 mL would be converted to 0.500 L.

Concentration Determination

Concentration refers to the amount of solute present in a given volume of solution. The most common unit for expressing concentration is molarity (M), defined as moles of solute per liter of solution. Calculating molarity involves dividing the moles of solute by the volume of the solution in liters.
For instance, if you have 0.0833 moles of glucose dissolved in 0.500 L of solution, the molarity is calculated as:\[ M = \frac{0.0833 \text{ moles}}{0.500 \text{ L}} = 0.167 \, \text{M} \] This means there are 0.167 moles of glucose per liter of solution. Understanding concentration is vital for predicting how a solution will behave in a chemical reaction and ensuring that solutions are prepared accurately.