Molarity and Solution Calculations

Molarity and Solution Calculations

Definition of Molarity

Molarity (M) is a measure of concentration that expresses the number of moles of solute per liter of solution. It is a fundamental concept in chemistry for preparing and analyzing solutions.

• Formula:

$\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Key Terms:

  • Solute: The substance dissolved in the solution.

  • Solution: The homogeneous mixture of solute and solvent.

Calculating Molarity

To calculate the molarity of a solution, divide the number of moles of solute by the total volume of the solution in liters.

• Example: Calculate the molarity of a solution prepared by dissolving 2.7 g NaOH in enough water to make 2.50 L of solution.

Step 1: Convert grams of NaOH to moles: $\text{moles NaOH} = \frac{2.7\ \text{g}}{40.00\ \text{g/mol}} = 0.0675\ \text{mol}$ Step 2: Calculate molarity: $M = \frac{0.0675\ \text{mol}}{2.50\ \text{L}} = 0.027\ \text{M}$

Using Molarity to Calculate Unknowns

Molarity can be used as a conversion factor to relate the amount of solute to the volume of solution. This is useful for determining unknown quantities in chemical reactions and laboratory preparations.

• Given Amount and Conversion Factors: Use the molarity as a conversion factor to relate moles and volume.

• Example: How many grams of Na3PO4 (MW: 163.94 g/mol) are present in 300.0 mL of a 0.500 M Na3PO4 solution?

Step 1: Convert volume to liters: $300.0\ \text{mL} = 0.3000\ \text{L}$ Step 2: Calculate moles: $0.3000\ \text{L} \times 0.500\ \text{mol/L} = 0.150\ \text{mol}$ Step 3: Convert moles to grams: $0.150\ \text{mol} \times 163.94\ \text{g/mol} = 24.6\ \text{g}$

Practice Problems

• Problem 1: What volume (in μL) of 0.125 M HBr contains 0.175 moles HBr?

• Solution: $1.40 \times 10^3\ \mu L$

• Problem 2: Hypernatremia is a medical condition where a patient has high sodium levels in their blood. If a patient has a sodium level of 165 mmol/L and 7.30 mL of blood is drawn, what mass (mg) of sodium would be present?

• Solution: $9.0\ \text{mg Na}^+$

• Problem 3: 2.44 grams of an unknown compound are dissolved in water to yield 100 mL of solution. The concentration is 0.675 M. What is the molecular weight of the substance?

• Solution: $36\ \text{g/mol}$

• Problem 4: A solution with a final volume of 750.0 mL was prepared by dissolving 30.00 mL of benzene (C6H6, density = 0.877 g/mL) in dichloromethane. Calculate the molarity of benzene in the solution.

• Solution: $1.14\ \text{mol/L}$

Summary Table: Molarity Calculations

Step	Description	Equation/Conversion
1	Convert mass to moles	$\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$
2	Convert volume to liters	$\text{volume (L)} = \frac{\text{volume (mL)}}{1000}$
3	Calculate molarity	$M = \frac{\text{moles}}{\text{liters}}$

Additional info: Molarity is widely used in stoichiometric calculations, titrations, and preparing laboratory solutions. It is essential for understanding solution chemistry and quantitative analysis.