Answered

Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

A 59.0 mL portion of a 1.80 M solution is diluted to a total volume of 258 mL. A 129 mL portion of that solution is diluted by adding 183 mL of water. What is the final concentration

Sagot :

maacastrobr

Answer:

0.170 M

Explanation:

As this is a series of dilutions, we can continuosly use the C₁V₁=C₂V₂ formula to solve this problem:

For the first step:

  • 59.0 mL * 1.80 M = 258 mL * C₂
  • C₂ = 0.412 M

Then for when 129 mL of that 0.412 M are diluted by adding 183 mL of water:

  • V₂ = 129 mL + 183 mL = 312 mL

Using C₁V₁=C₂V₂:

  • 129 mL * 0.412 M = 312 mL * C₂
  • C₂ = 0.170 M
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.