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motikmotik
3 years ago
10

Practically: Add 1.66 ml of my 0.3M lemonade to a 15 ml microcentrifuge tube. Add 3.33 ml of your diluent (water, in this case)

to bring your volume up to 5 ml. Mix your solution either by vortexing or by pipetting up and down with your pipetteman. Let's make sure you got this concept by answering a question below. I have made 15 ml of 200 mM CaCl2 stock and need to make 40 ml of 50mM for my experiment. How much of my concentrated stock solution (in milliliters) and how much water do I need to mix to make the 40 ml of 50mM CaCl2
Chemistry
1 answer:
Sedbober [7]3 years ago
4 0

Answer:

We would need 10 mL of the concentrated CaCl₂ stock solution, and 30 mL of water.

Explanation:

To solve the question asked we can use the C₁V₁=C₂V₂ equation, where:

  • C₁ = 200 mM
  • V₁ = ?
  • C₂ = 50 mM
  • V₂ = 40 mL

We <u>solve for V₁</u>:

  • V₁ = 10 mL

We would need 10 mL of the concentrated CaCl₂ stock solution, and (40-10) 30 mL of water.

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yield = 52.23 %

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mass of aluminium = mass of bottle with aluminium pieces - bottle mass

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number of moles of aluminium = 0.9974 / 27 = 0.03694 moles

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To calculate the theoretical quantity of alum that should be obtained from 0.03694 moles of aluminium we devise the following reasoning:

if       2 moles of aluminium produce 2 moles of alum

then 0.03694 moles of aluminium produce X moles of alum

X = (0.03694 × 2) / 2 = 0.03694 moles of alum (theoretical)

yield = (practical quantity / theoretical quantity) × 100

yield = (0.01929 /  0.03694) × 100

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