How do I calculate the molarity of the Na2CO3 solution? My teacher isn't very good at helping me solve this problem. I keep getting the answer, but I don't know why I am getting it. Could you please help me understand this problem and why you should do this specific step?


  A $\pu{1.00 g}$ sample of $\ce{Na2CO3.10H2O}$ was dissolved in $\pu{20.0 mL}$ of distilled water. Additional water was added so as to give $\pu{250 mL}$ of solution. What is the molar concentration of $\ce{Na2CO3}$?


Here is what I did and got the answer and don't understand why I need to do it:

\begin{align}
  \frac{\pu{1 g}~\ce{Na2CO3.10H2O}}{\pu{286 g}~\ce{Na2CO3.10H2O}} &amp;= \pu{0.003447 mol}\\
\frac{\pu{0.003447 mol}}{\pu{0.020 L}} &amp;= \pu{0.174825 M}\\
(\pu{0.174825 M})(\pu{0.020 L}) &amp;= (\pu{0.250 L})(x)\\
x &amp;= \pu{0.013986 M}
\end{align}

$\pu{0.013986 M}$ is the correct answer, but I don't know why and don't understand the question.
If you were to do the problem, could you explain why you did what you did to get to the answer.
The first part makes more sense writing moles explicitly. The first step is determining how many moles of sodium carbonate are in the original $1.00\ \mathrm  g$ sample:

$$\frac{1.00\ \mathrm{g}}{286\ \mathrm{g/mol}} = 0.003447\ \mathrm{mol}$$

The second step is determining the concentration of the first $20.0\ \mathrm{mL}$ solution, i.e. what concentration results from dissolving the $0.00345\ \mathrm{mol}$ of sodium carbonate calculated above.
$$\frac{0.003447\ \mathrm{mol}}{0.0200\ \mathrm{L}} = 0.174825\ \mathrm{M}$$

The third part is simply a dilution – what concentration results when we dilute the first $20.0\ \mathrm{mL}$ solution to $250\ \mathrm{mL}$:
$$c_2 = \frac{c_1V_1}{V_2}$$
$$c_2 = \frac{(0.174825\ \mathrm{M})(0.020\ \mathrm{L})}{(0.250\ \mathrm{L})} = 0.0140\ \mathrm{M}$$