Solution Concentration Fundamentals
📌 Key concentration expressions covered include mass percent, volume percent, mole fraction, molality (m), and molarity (M).
💧 The fundamental relationship is: Solute + Solvent = Solution.
🧂 In a binary mixture, the solute is the substance being dissolved, and the solvent is the substance doing the dissolving (e.g., NaCl is solute, Water is solvent).
⚖️ Density is defined as $\text{Density} = \text{Mass} / \text{Volume}$, often expressed in $\text{g/mL}$ or $\text{g/cm}^3$.

Formulas for Concentration Calculations
📏 Mass Percent: $\frac{\text{Mass of Solute}}{\text{Mass of Solution}} \times 100\%$
🧪 Volume Percent ($v$%): $\frac{\text{Volume of Solute}}{\text{Volume of Solution}} \times 100\%$
➗ Mole Fraction ($\chi$) of substance A: $\frac{\text{Moles of A}}{\text{Total Moles}} = \frac{n_A}{n_A + n_B + ...}$
⚡ Molarity (M): $\frac{\text{Moles of Solute}}{\text{Liters of Solution}}$
⚖️ Molality (m): $\frac{\text{Moles of Solute}}{\text{Kilograms of Solvent}}$ (Crucial difference: uses mass of solvent, not solution).

Worked Examples and Application
🥇 Example 1 (Mass Percent): 15 g NaCl in 225 g water resulted in a mass percent of $\mathbf{6.25\%}$ (calculated as $15 \text{ g} / (15 \text{ g} + 225 \text{ g}) \times 100\%$).
🧪 Example 2 (Volume Percent): 25 mL methanol in 150 mL water yielded a volume percent of methanol of $\mathbf{14.28\%}$ (calculated as $25 \text{ mL} / 175 \text{ mL} \times 100\%$).
💧 Example 2 (Mass Percent of Water): Required using densities to convert volumes to masses, resulting in $\mathbf{88.3\%}$ mass percent of water.
⚛️ Example 4 (Mole Fraction): Calculating the mole fraction of NaF required first converting 25 g NaF (Molar Mass $\approx 42 \text{ g/mol}$) and 200 g H₂O (Molar Mass $\approx 18.016 \text{ g/mol}$) into moles. The resulting mole fraction was $\mathbf{0.051}$.
🔥 Example 7 (Complex Calculation): Converting a $\mathbf{15\%}$ mass percent solution of $\text{AlCl}_3$ with a density of $1.17 \text{ g/mL}$ to molarity involved assuming $100 \text{ g}$ of solution, calculating moles of solute ($\text{AlCl}_3$, MM $\approx 133.33 \text{ g/mol}$), and converting the solution mass to liters ($\mathbf{0.08547 \text{ L}}$), yielding a molarity of $\mathbf{1.316 \text{ M}}$.

Key Points & Insights
➡️ Always distinguish clearly between Molality (kg of solvent) and Molarity (L of solution) as they share the same numerator (moles of solute).
➡️ When solving problems involving density conversions, ensure volume units match: $1 \text{ L} = 1000 \text{ mL}$.
➡️ For volume/mass percent involving two liquids (like methanol and water), the solute is typically the component present in the lesser quantity, unless the question explicitly dictates which substance to calculate the percentage for.
➡️ When given mass percent and density, use the mass percent to establish a ratio based on $100 \text{ g}$ of solution to find the required mass of solute, and then use density to convert the solution mass into the required volume (Liters for Molarity).

📸 Video summarized with SummaryTube.com on Dec 26, 2025, 06:45 UTC