Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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 :

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
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.