# Calculating the volume of 2 M solution needed to make a 0.1 M solution

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$$\pu{100 ml}$$?

• 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. – Todd Minehardt May 21 at 17:31

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 $$\pu{100 mL}$$.