A science educator's experiment requires 300 mL of a 15% salt solution. She has 10% and 25% solutions. How many mL of the 25% solution are needed?

["Title: How a Science Educator Solved a Salt Solution Problem: Mixing 10% and 25% Solutions", "When preparing a precise salt solution, science educators often rely on accurate dilution calculations. Recently, a dedicated science educator faced a practical challenge: preparing exactly 300 mL of a 15% salt solution using only 10% and 25% salt solutions. This hands-on experiment not only tests knowledge of concentration mixing but also reflects real-world applications in education and laboratory work. Let’s explore how this problem is solved using basic algebra and the concept of mass balance.", "---", "### The Zeroth Concept: Mixing Solutions by Concentration", "To create 300 mL of a 15% salt solution, we mix two stock solutions:\n- 10% sodium chloride\n- 25% sodium chloride", "The total volume must equal 300 mL, and the amount of salt (in grams) from both solutions must add up to 15% of 300 mL.", "---", "### Step 1: Define the Variables", "Let:\n- ( x ) = volume (in mL) of the 25% solution needed\n- Then, ( 300 - x ) = volume (in mL) of the 10% solution needed", "---", "### Step 2: Set Up the Equation Based on Total Salt", "Salt amount from each solution must sum to the desired amount in the final mixture:", "[\n\ ext{Salt from 25% solution} + \ ext{Salt from 10% solution} = \ ext{Total salt in final solution}\n]", "In grams, this translates to:", "[\n0.25x + 0.10(300 - x) = 0.15 \ imes 300\n]", "---", "### Step 3: Solve the Equation", "Calculate the right-hand side:", "[\n0.15 \ imes 300 = 45 \ ext{ grams of salt}\n]", "Now expand the left-hand side:", "[\n0.25x + 0.10 \ imes 300 - 0.10x = 45\n]\n[\n0.25x + 30 - 0.10x = 45\n]\n[\n(0.25 - 0.10)x + 30 = 45\n]\n[\n0.15x + 30 = 45\n]\n[\n0.15x = 15\n]\n[\nx = \frac{15}{0.15} = 100\n]", "---", "### Step 4: Interpret the Result", "The educator needs 100 mL of the 25% salt solution. This means 200 mL of the 10% solution is also required, since ( 300 - 100 = 200 ).", "Check the salt balance:\n- Salt from 25% solution: ( 0.25 \ imes 100 = 25 ) g\n- Salt from 10% solution: ( 0.10 \ imes 200 = 20 ) g\n- Total salt: ( 25 + 20 = 45 ) g\n- Final concentration: ( \frac{45}{300} = 15% ) — correct!", "---", "### Why This Experiment Matters", "This simple yet precise mixing problem is a staple in science education. It teaches important principles like:\n- Conservation of mass in solutions\n- Linear equations applied to real-life scenarios\n- The importance of accuracy in laboratory work", "By solving how much of each stock solution is needed, students and educators alike deepen their understanding of stoichiometry, dilutions, and practical chemistry.", "---", "### Final Answer", "To prepare 300 mL of a 15% salt solution using 10% and 25% stock solutions, the educator must measure 100 mL of the 25% solution and 200 mL of the 10% solution.", "This experiment exemplifies how science educators use math and chemistry to achieve precise, educational outcomes — one drop at a time.", "---", "Keywords: salt solution experiment, concentration mixing, 10% and 25% salt solutions, algebra in chemistry, science educator, solution preparation, stoichiometry, dilution calculation, STEM education.\nMeta Description: Learn how a science educator accurately mixes a 15% salt solution by combining 10% and 25% solutions using a simple algebra approach. Perfect for chemistry students and teachers."]