# Question 2ade7

Nov 23, 2016

${\text{0.48 mol L}}^{- 1}$

#### Explanation:

The thing to remember about dilution calculations is that the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.

You thus have

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{V}_{\text{diluted"/V_"concentrated" = c_"concentrated"/c_"diluted}}} |}}}$

These two ratios give you the dilution factor.

In your case, the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is

V_"diluted"/V_"concentrated" = (400. color(red)(cancel(color(black)("L"))))/(10. color(red)(cancel(color(black)("L")))) = 40

This means that the concentration of the concentrated solution is $40$ times higher than the concentration of the diluted solution.

${c}_{\text{concnetrated"/c_"diluted}} = 40$

Therefore, you have

${c}_{\text{diluted" = c_"concentrated}} / 40$

In your case, this will be equal to

c_"diluted" = ("19.1 mol L"^(-1))/40 = color(darkgreen)(ul(color(black)("0.48 mol L"^(-1))))#

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the concentrated solution.