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Mathematics
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Wow! That's it?!

OpenStudy (anonymous):

I think so, what's 1.0%(w/v) is that the same as Molarity?

OpenStudy (anonymous):

It stands for (mass/volume)= [Mass of solute (g) / Volume of solution (ml)] x 100

OpenStudy (anonymous):

Molarity is moles/liter

OpenStudy (anonymous):

Hmmm, this just got tougher.

OpenStudy (anonymous):

Haha, yes. It sucks.

OpenStudy (anonymous):

is w/v weight/volume?

OpenStudy (anonymous):

Ya :)

OpenStudy (anonymous):

How does the 1.0% come in? Is it 1g/L?

OpenStudy (anonymous):

1% apparently = 10mg/ml

OpenStudy (anonymous):

Is that 2.0 mols of NaOH or 2.0 Molarity of NaOH?

OpenStudy (anonymous):

It's 2.0 molarity (mol/liter)

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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!

OpenStudy (anonymous):

Woops, should be\[1ml/10mg\]

OpenStudy (anonymous):

\[7.9996NaOH\]Other than that I'm stumped.

OpenStudy (anonymous):

That actually sounds like a reasonable answer. Thank you!!

OpenStudy (anonymous):

You're welcome :)

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