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anonymous

  • 5 years ago

Describe how to prepare 1.00 L of 1.0% (w/v) NaOH from a 2.0 M NaOH stock

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  1. anonymous
    • 5 years ago
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    \[V_1M_1=V_2M_2\]\[(1.00L)(1.0M)=V_2(2.0M)\]\[1.00mol=V_2(2.0M)\]/2.0M /2.0M\[.5L=V_2\]Chemistrty :D

  2. anonymous
    • 5 years ago
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    Wow! That's it?!

  3. anonymous
    • 5 years ago
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    I think so, what's 1.0%(w/v) is that the same as Molarity?

  4. anonymous
    • 5 years ago
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    It stands for (mass/volume)= [Mass of solute (g) / Volume of solution (ml)] x 100

  5. anonymous
    • 5 years ago
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    Molarity is moles/liter

  6. anonymous
    • 5 years ago
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    Hmmm, this just got tougher.

  7. anonymous
    • 5 years ago
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    Haha, yes. It sucks.

  8. anonymous
    • 5 years ago
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    is w/v weight/volume?

  9. anonymous
    • 5 years ago
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    Ya :)

  10. anonymous
    • 5 years ago
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    How does the 1.0% come in? Is it 1g/L?

  11. anonymous
    • 5 years ago
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    1% apparently = 10mg/ml

  12. anonymous
    • 5 years ago
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    Is that 2.0 mols of NaOH or 2.0 Molarity of NaOH?

  13. anonymous
    • 5 years ago
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    It's 2.0 molarity (mol/liter)

  14. anonymous
    • 5 years ago
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    I got\[2.0molNaOH/1L \times39.998g NaOH/1mol \times 1000mg/1g \times 1ml/1mg \times 1L/1000ml\]All my units cancel, and I'm left with\[79.996\]So now I'm confused.

  15. anonymous
    • 5 years ago
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    Haha, this looks like my answer. I am going on two hours of trying to figure this one out. I think I will just combine our answers and turn that in :) Thanks for all of your help!!! I really, really appreciate it!

  16. anonymous
    • 5 years ago
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    Woops, should be\[1ml/10mg\]

  17. anonymous
    • 5 years ago
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    \[7.9996NaOH\]Other than that I'm stumped.

  18. anonymous
    • 5 years ago
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    That actually sounds like a reasonable answer. Thank you!!

  19. anonymous
    • 5 years ago
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    You're welcome :)

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