If I have a 1M solution of NaOH and I want to obtain a 0.2M solution. Then according to the formula $$ {M_1V_1 = M_2V_2}$$ I would get the answer as 200 mL. In my textbook they have said that "Thus, 200 mL of 1M NaOH is taken and enough water is added to dilute it to make it 1 liter. Note that the number of moles of solute was 0.2 in 200 mL and it has remained same , i.e., 0.2 even after dilution as we have just changed just the amount of solvent and have not done anything with respect to NaOH. But keep in mind the concentration.". So my question is, by adding water to the 0.2M solution to make it one liter wouldn't the molarity change?

  • $\begingroup$ You have reversed the steps. You have answered your question first and then you have asked the question. $\endgroup$
    – Poutnik
    Jun 1, 2019 at 7:48
  • $\begingroup$ The answer was in my textbook and I'm unable to understand it $\endgroup$
    – user662650
    Jun 1, 2019 at 7:49
  • $\begingroup$ What have you done to understand it ? It is based on very trivial math of a basic school. $\endgroup$
    – Poutnik
    Jun 1, 2019 at 7:54
  • $\begingroup$ I have so far understood that 200 mL of a 1M solution gives me a 0.2M. But if we add water to make it one litre then won't the molarity change? $\endgroup$
    – user662650
    Jun 1, 2019 at 7:56
  • 2
    $\begingroup$ This question shows one reason why the quantity "concentration" should not be called "molarity". $\endgroup$
    – user7951
    Jun 1, 2019 at 14:30

1 Answer 1


I think you have confused amount of substance with molarity of the solution. For example:

  • $\ce{NaOH}$ solution with $\pu{1M}$ in molarity means $\pu{1L}$ of solution contains $\pu{1mol}$ of $\ce{NaOH}$.
  • You can calculate amount of $\ce{NaOH}$ in any volume of the solution by multiplying its molarity and required volume in $\pu{L}$: $$\text{Amount of }\ce{NaOH}\text{ in } \pu{mol} = \text{molarity of the solution} \times \text{volume in }\pu{L} = MV$$

Now see what happens when you measured $\pu{200 mL}$ of $\pu{1M} \; \ce{NaOH}$ solution: $$\text{The amount of }\ce{NaOH}\text{ in } \pu{mol} = \text{molarity of the solution} \times \text{volume in }\pu{L}$$ $$= \pu{1 mol\:L^{-1}} \times \pu{200 mL} \times \frac{\pu{1L}}{\pu{1000 mL}}=\pu{0.2 mol}$$

If you diluted this $\pu{200 mL}$ volume to $\pu{1L}$ what happens to its molarity?

That solution still contains $\pu{0.2 mol}$ of $\ce{NaOH}$, because you add nothing but water.

Now, $\pu{1 L}$ of new solution contain $\pu{0.2 mol}$ of $\ce{NaOH}$. Thus, by definition (see the first bullet above), its molarity is $\pu{0.2 M}$ (or $\pu{0.2 mol\:L^{-1}}$).


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