Answered

Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive 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 :

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
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.