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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