# 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
• Please visit this page, this page and this ‎one on how to format your posts better.‎ Alternatively, visit this chatroom for further formatting guidance. – M.A.R. Sep 6 '15 at 17:14
• 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