Describe how to prepare 1.00 L of 1.0% (w/v) NaOH from a 2.0 M NaOH stock
\[V_1M_1=V_2M_2\]\[(1.00L)(1.0M)=V_2(2.0M)\]\[1.00mol=V_2(2.0M)\]/2.0M /2.0M\[.5L=V_2\]Chemistrty :D
Wow! That's it?!
I think so, what's 1.0%(w/v) is that the same as Molarity?
It stands for (mass/volume)= [Mass of solute (g) / Volume of solution (ml)] x 100
Molarity is moles/liter
Hmmm, this just got tougher.
Haha, yes. It sucks.
is w/v weight/volume?
How does the 1.0% come in? Is it 1g/L?
1% apparently = 10mg/ml
Is that 2.0 mols of NaOH or 2.0 Molarity of NaOH?
It's 2.0 molarity (mol/liter)
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.
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!
Woops, should be\[1ml/10mg\]
\[7.9996NaOH\]Other than that I'm stumped.
That actually sounds like a reasonable answer. Thank you!!
You're welcome :)
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