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A research lab has 150 test tubes, each containing a solution at a concentration of 0.5 M. If 20% of the solution is removed and replaced with a 1.5 M solution, what is the final concentration of the mixture in one test tube?
This scenario reflects a common procedure in scientific workflows: updating solutions by swapping portions to test reactivity, concentration stability, or formulation models—practices relevant in labs exploring chemical interactions, drug development, or material science. For curious learners and professionals in the U.S. research community, understanding these precise dilution dynamics is key to replicating results and driving innovation safely. Why A research lab has 150 test tubes, each containing a solution at a concentration of 0.5 M. If 20% of the solution is removed and replaced with a 1.5 M solution, what is the final concentration of the mixture in one test tube?
This question highlights a fundamental concept in solution chemistry: mixing a sample fraction with a stronger solution triggers a new equilibrium concentration based on volume and molarity. The lab setting underscores precision—critical when working with reagents where small concentration shifts impact experimental outcomes. With 150 test tubes involved, consistency across each becomes a baseline for reliable data collection and peer validation. How A research lab has 150 test tubes, each containing a solution at a concentration of 0.5 M. If 20% of the solution is removed and replaced with a 1.5 M solution, what is the final concentration of the mixture in one test tube?
The process unfolds by removing 20% of the original 0.5 M solution, leaving 80% at 0.5 M. From this remainder, 20% volume becomes fresh 1.5 M solution. The final concentration blends both sources proportionally by moles:
- Moles from original: 0.8 × 0.5 M × 150 tubes = 60 moles
- Moles from new: 0.2 × 150 tubes × 1.5 M = 45 moles
Total moles = 60 + 45 = 105 moles
Total volume = 150 tubes × 1.0 L = 150 L
Final concentration = 105 moles / 150 L = 0.7 M