# What is the concentration of the solution produced when 150.0 mL of water is added to 200.0 mL of 0.250 M NaCl?

Feb 13, 2018

$\text{0.143 M}$

#### Explanation:

The thing to remember about a dilution is that you can find its dilution factor by dividing either the volume of the diluted solution by the volume of the stock solution

$\text{DF" = V_"diluted"/V_"stock}$

or by dividing the concentration of the stock solution by the concentration of the diluted solution.

$\text{DF" = c_"stock"/c_"diluted}$

So you can say that for any dilution, you have

$\text{DF" = V_"diluted"/V_"stock" = c_"stock"/c_"diluted}$

In your case, the volume of the diluted solution will be

${V}_{\text{diluted" = "150.0 mL + 200.0 mL" = "350.0 mL}}$

This means that the dilution factor is equal to

"DF" = (350.0 color(red)(cancel(color(black)("mL"))))/(200.0 color(red)(cancel(color(black)("mL")))) = color(blue)(1.75)

You can thus say that the stock solution was $\textcolor{b l u e}{1.75}$ times more concentrated than the diluted solution, since

${c}_{\text{stock" = "DF" * c_"diluted}}$

This implies that the concentration of the diluted solution is equal to

${c}_{\text{diluted" = c_"stock"/"DF}}$

c_"stock" = "0.250 M"/color(blue)(1.75) = color(darkgreen)(ul(color(black)("0.143 M")))

The answer is rounded to three sig figs.