# Concentration calculation?

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

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

$\mathrm{V_1 = \frac{0.1~M \times 100~ml}{2~M} = 5~ml}$??

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

• Please don't post text as images because they are not searchable by future users. – bon Sep 6 '15 at 17:13
• This answer seems correct. What are you confused about? – bon Sep 6 '15 at 17:22
• @bon I dunno, I just get really frazzled when I have to do solutions and concentrations. Thanks. – Ro Siv Sep 6 '15 at 17:30

If you have a simple solution of say 0.1 M sodium chloride, you have 0.1M NaCl in 100ml. Which is equivalent to "100 mL of 0.1 moles NaCl per litre", Which is equal to ((100/1000) * 0.1). which is equal to 0.01 moles of NaCl.

5 mL of 2M NaCl is equivalent to "5 mL of 2 moles NaCl per litre" Which is equal to ((5/1000) * 2). which is equal to 0.01 moles of NaCl.

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 - 2M is 20x more concentrated than 0.1 (2/0.1), so you expect to need 20x less of the 2M than the 0.1M.

5 ml is 20x less than 100 mL