A chemistry student prepares a solution by mixing 150 mL of a 20% salt solution with 250 mL of a 40% salt solution. What is the concentration of salt in the resulting mixture?

A chemistry student prepares a solution by mixing 150 mL of a 20% salt solution with 250 mL of a 40% salt solution. What is the concentration of salt in the resulting mixture?

["Chemistry Student Solution Preparation: Calculating Salt Concentration in a Mixed Solution", "Mixing solutions of different concentrations is a fundamental process in chemistry, especially in practical laboratory work. A common question arises when combining saltwater solutions: What is the concentration of salt in the resulting mixture? This article walks through the calculation step-by-step for a chemistry student preparing a solution by mixing 150 mL of a 20% salt solution with 250 mL of a 40% salt solution. By applying principles of mass and volume concentration, we determine the final concentration of salt in the combined mixture.", "### The Problem: Mixing Two Salt Solutions", "Consider two solutions:", "- Solution A:\n - Volume: 150 mL\n - Salt concentration: 20%", "- Solution B:\n - Volume: 250 mL\n - Salt concentration: 40%", "We need to compute the concentration of salt in the combined solution after mixing.", "---", "### Step 1: Calculate Salt Mass in Each Solution", "Since concentration is expressed as a percentage by mass/volume, we calculate the amount of salt using:", "[\n\ ext{Mass of salt} = \ ext{Volume} \ imes \ ext{Concentration}\n]", "- For Solution A:\n[\n150~\ ext{mL} \ imes 0.20 = 30~\ ext{grams of salt}\n]", "- For Solution B:\n[\n250~\ ext{mL} \ imes 0.40 = 100~\ ext{grams of salt}\n]", "---", "### Step 2: Total Salt and Total Volume", "Add the salt masses and volumes:", "- Total salt:\n[\n30~\ ext{g} + 100~\ ext{g} = 130~\ ext{grams of salt}\n]", "- Total volume:\n[\n150~\ ext{mL} + 250~\ ext{mL} = 400~\ ext{mL}\n]", "---", "### Step 3: Calculate Final Concentration", "Concentration is defined as (mass of solute / volume of solution) × 100%:", "[\n\ ext{Final concentration} = \left( \frac{130~\ ext{g}}{400~\ ext{mL}} \right) \ imes 100% = 32.5%\n]", "---", "### Final Answer", "The resulting mixture has a salt concentration of 32.5%. This calculation demonstrates the principle of conservation of mass in solution mixing and is essential for students learning dilution, molarity, and analytical techniques in chemistry.", "---", "Understanding such calculations not only supports academic success but also prepares students for real-world lab work where precise salt solutions are used in experiments, monitoring, and quality tests. Mastering these concepts ensures accuracy and confidence in scientific reasoning and practical application."]

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