Solutions and Concentration Calculations: Study Notes for Introductory Chemistry

Solutions and Solution Concentration

Speeding Up the Dissolving Process

Several factors can influence the rate at which a solute dissolves in a solvent. Understanding these can help in laboratory and everyday settings.

• Increasing Temperature: When the solution is warmed, particles move faster, increasing the rate of dissolution.

• Stirring the Solution: Stirring spreads solute particles throughout the solvent, increasing contact and speeding up dissolution.

• Crushing the Solute: Breaking solute into smaller pieces increases surface area exposed to the solvent, which increases the dissolving rate.

Example: Dissolving a sugar cube in water is faster if the water is warm, the cube is crushed, and the solution is stirred.

Calculating Solution Concentration (Percent by Mass)

The concentration of a solution can be expressed as percent by mass, which is the mass of solute divided by the total mass of the solution, multiplied by 100.

• Formula: $\text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

• Example Calculation: Dissolving 8.0 g NaCl in 100 g water: $\frac{8.0}{108} \times 100 = 7.41\%$

Calculating Mass of Solute in a Given Solution

To find the mass of solute in a solution of known concentration and volume:

• Step 1: Calculate total mass of solution.

• Step 2: Use percent concentration to find mass of solute.

• Example: In a 7.5% glucose solution, 25 g glucose is present in every 100 g solution. For 275 g solution: $\frac{25}{100} \times 275 = 68.75$ g glucose.

Molarity (M) and Solution Calculations

Molarity is a common unit of concentration, defined as moles of solute per liter of solution.

• Formula: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Example: To prepare 0.625 M CuSO4 solution, calculate moles and mass needed for a given volume.

Steps for Preparing Solutions of Known Molarity

1. Calculate moles of solute needed: $\text{moles} = M \times \text{volume (L)}$

2. Convert moles to grams using molar mass: $\text{mass} = \text{moles} \times \text{molar mass}$

3. Dissolve calculated mass in appropriate volume of solvent.

Example: To make 125 mL of 0.625 M CuSO4 solution:

• Volume = 0.125 L

• Moles needed = $0.625 \times 0.125 = 0.078125$ mol

• Molar mass of CuSO4 = 63.5 + 32 + (16 × 4) = 159.5 g/mol

• Mass needed = $0.078125 \times 159.5 = 12.47$ g

Dilution Calculations

When diluting a solution, the amount of solute remains constant, but the volume increases, lowering the concentration.

• Formula: $M_1 V_1 = M_2 V_2$ Where $M_1$ and $V_1$ are initial molarity and volume, $M_2$ and $V_2$ are final molarity and volume.

• Example: To dilute 175 mL of 0.25 M HCl to 0.10 M, solve for final volume: $V_2 = \frac{M_1 V_1}{M_2} = \frac{0.25 \times 175}{0.10} = 437.5$ mL

Sample Table: Solution Preparation and Dilution

Step	Calculation	Result
Moles of solute	$M \times V$	e.g., $0.625 \times 0.125 = 0.078125$ mol
Mass of solute	Moles × Molar mass	e.g., $0.078125 \times 159.5 = 12.47$ g
Dilution volume	$V_2 = \frac{M_1 V_1}{M_2}$	e.g., $\frac{0.25 \times 175}{0.10} = 437.5$ mL

Additional info:

• These calculations are foundational for laboratory work, chemical manufacturing, and medical applications.

• Always use proper units and significant figures in calculations.