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I need help with a basic concentration problem.

If I have a $\mathrm{2~M}$ solution, and want to make a $\mathrm{0.1~M}$ solution of $\mathrm{100~ml}$, how would I go about that?

My work:

$$M_1V_1 = M_2V_2$$

Where $M_1 = \text{initial concentration}$, $M_2 = \text{final concentration}$, $V_1 = \text{volume needed from initial}$, $V_2 = \text{volume needed from final}$

$$2~M \times V_1 = \pu{0.1 M \times 100 ml}$$

$$V_1 = \frac{\pu{0.01 M\cdot L}}{\pu{2~M}} = \pu{5 ml}?$$

Therefore I need $\pu{5 ml}$ of my $\pu{2 M}$ solution, increased to $\pu{100 ml}$ to make a $\pu{0.1 M}$ solution of$\pu{100 ml}$?

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  • $\begingroup$ Please note that I have amended your post in one area where the units were incorrect; and I have explicitly used liters, because molarity is in units of moles per liter. $\endgroup$ – Todd Minehardt May 21 at 17:31
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If you have a simple solution of say $\pu{0.1 M}$ sodium chloride, you have $\pu{0.1M}\, \ce{NaCl}$ in $\pu{100 ml}$.

Which is equivalent to "$\pu{100 mL}$ of $\pu{0.1 mol}\,\ce{NaCl}$ per litre", or,

$$n_\ce{NaCl} =\frac{100}{1000} \times 0.1 = \pu{0.01 mol}$$

$\pu{5 mL}$ of $\pu{2 M}\ \ce{NaCl}$ is equivalent to "$\pu{5 mL}$ of $\pu{2 mol}$ $\ce{NaCl}$ per litre", or,

$$n_\ce{NaCl} = \frac{5}{1000} \times 2 = \pu{0.01 mol}$$

So congratulations, you did it right. If ever you aren't sure, just write out all the steps and follow it through.

You could also look at it like this - $\pu{2 M}$ is $20\times$ more concentrated than $0.1\times\frac{2}{0.1}$, so you expect to need $20\times$ less of the $\pu{2 M}$ than the $\pu{0.1 M}$.

$\pu{5 ml}$ is $20\times$ less than $\pu{100 mL}$.

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