#### \(\frac{7}{3}\)1. A laboratory experiment requires a solution with 150 mL of a 20% saline solution. The technician has a 30% saline solution and a 10% saline solution available. How many mL of each solution should be mixed to achieve the desired concentration?
["How to Prepare 150 mL of 20% Saline Solution Using 30% and 10% Solutions: A Practical Lab Experiment", "Preparing accurate saline solutions is a fundamental task in laboratory settings. In this experiment, the objective is to obtain exactly 150 mL of a 20% saline solution by mixing two available concentrations: a 30% saline solution and a 10% saline solution. This article explains the step-by-step calculation using algebraic reasoning to determine the correct volumes needed—saving time, reducing waste, and ensuring precision in experimental setups.", "---", "### Understanding the Problem", "We need a final solution of:", "- Total volume: 150 mL\n- Final concentration: 20% saline\n- Available stock solutions:\n - 30% saline solution (stronger)\n - 10% saline solution (weaker)", "We are to find:\n- How many mL of the 30% solution to use\n- How many mL of the 10% solution to use", "---", "### Step 1: Define variables", "Let:\n- ( x ) = volume (in mL) of the 30% saline solution\n- ( y ) = volume (in mL) of the 10% saline solution", "From the total volume constraint:", "[\nx + y = 150 \quad \ ext{(Equation 1)}\n]", "From the total mass (salt) requirement:", "The final solution must contain ( 20% \ imes 150 = 30 ) g of salt (assuming 1 mL ≈ 1 g in dilute solutions for approximation).", "The total salt from both solutions must equal 30 g:", "[\n0.30x + 0.10y = 30 \quad \ ext{(Equation 2)}\n]", "---", "### Step 2: Solve the system of equations", "From Equation 1:", "[\ny = 150 - x\n]", "Substitute into Equation 2:", "[\n0.30x + 0.10(150 - x) = 30\n]", "Expand:", "[\n0.30x + 15 - 0.10x = 30\n]", "Combine like terms:", "[\n0.20x + 15 = 30\n]", "Subtract 15 from both sides:", "[\n0.20x = 15\n]", "Divide by 0.20:", "[\nx = \frac{15}{0.20} = 75\n]", "Now, substitute back to find ( y ):", "[\ny = 150 - 75 = 75\n]", "---", "### Step 3: Verify the result", "- Volume of 30% solution: 75 mL → contributes ( 0.30 \ imes 75 = 22.5 ) g salt\n- Volume of 10% solution: 75 mL → contributes ( 0.10 \ imes 75 = 7.5 ) g salt\n- Total salt: ( 22.5 + 7.5 = 30 ) g\n- Total volume: 150 mL → concentration = ( \frac{30}{150} = 20% ), which matches\n✅ Both conditions satisfied!", "---", "### Conclusion", "To prepare 150 mL of a 20% saline solution using a 30% and a 10% stock solution, mix:", "- 75 mL of the 30% saline solution\n- 75 mL of the 10% saline solution", "This precise mixture ensures accurate laboratory results while minimizing trial and error.", "---", "### Why This Calculation Matters", "Accurate dilution calculations are essential in chemistry, medicine, and biomedical research to maintain experimental validity, ensure safety, and comply with protocols. Understanding how to set up and solve such equations helps technicians and researchers perform reliable, reproducible work quickly.", "---", "Keywords: saline solution preparation, lab experiment, concentration calculation, 30% saline, 10% saline, mixture of solutions, lab technique, stoichiometry, chemistry lab", "Meta Description:\nLearn how to mix 30% and 10% saline solutions to create 150 mL of a precise 20% saline solution using algebra. Solve volume and concentration problems efficiently."]
