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

🔗 Related Articles You Might Like:

📰 Solution: Treat the two letters as a single entity. This gives $4!$ arrangements for the entities (the letter pair and 3 manuscripts). The letters within the pair can be ordered in $2!$ ways. Total favorable arrangements: $4! imes 2!$. Total possible arrangements: $5!$. The probability is $ rac{4! imes 2!}{5!} = rac{2}{5}$. $oxed{\dfrac{2}{5}}$ 📰 Question: If a researcher studies 4 random integers between 0 and 10 inclusive, what is the probability that all are even? 📰 Solution: There are 11 integers (0–10), of which 6 are even (0, 2, 4, 6, 8, 10). The number of ways to choose 4 distinct even integers is $inom{6}{4}$. The total number of ways to choose 4 distinct integers is $inom{11}{4}$. The probability is $ rac{inom{6}{4}}{inom{11}{4}} = rac{15}{330} = rac{1}{22}$. $oxed{\dfrac{1}{22}}$ 📰 Seaport Diner 3414737 📰 Penguin Solutions That Will Revolutionize Your Business Overnight 8129779 📰 Price Club 6119519 📰 The Red Pony 3693281 📰 A P Left1 Fracrnrightnt 1000 Left1 Frac00452Right2 Times 3 1000 Times Left10225Right6 6700410 📰 First Hour Production 250 Widgets 5300983 📰 Alcaraz Vs Shelton 5604951 📰 All Flash Characters 5785732 📰 I Love You En Italiano 9203009 📰 Chinese Food Delivery Near Me 1862134 📰 Sonic The Hedgehog Before 2011 The Rad Test Area That Changed Everything 2113572 📰 You Wont Believe How Addictive The Color Game Isplay For Instant Wins 1425658 📰 No More Wired Nightmaresheres How To Put Your Printer Online Fast 2330776 📰 You Wont Believe What Happened In Nightmare On Elm Street 2Watch Before You Fall Asleep 4615120 📰 San Gennaro Festival 8435348