Question

What is the volume of `2.0` `mol` `dm^-3` sodium chloride needed to prepare `250cm^3` of `0.2``mol` `dm^-3` sodium chloride solution ?

What are the steps to solve this ?

Answer 1

`"25 cm"^3`

Explanation:

Start by calculating the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution.

`"DF" = (2.0 color(red)(cancel(color(black)("mol dm"^(-3)))))/(0.2color(red)(cancel(color(black)("mol dm"^(-3))))) = color(blue)(10)`

This ratio gives you the dilution factor, `"DF"`.

Now, the thing to remember about the dilution factor is that can also be calculated by taking the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

`"DF" = V_"diluted"/V_"stock"`

In your case, the diluted solution has a volume of `"250 cm"^(3)`, which means that the stock solution must have a volume of

`V_"stock" = "250 cm"^3/color(blue)(10) = color(darkgreen)(ul(color(black)("25 cm"^3)))`

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the concentration of the diluted solution.

So, in order to prepare this solution, you need to take `"25 cm"^3` of the `"2.0-mol dm"^(-3)` sodium chloride solution and add enough water to get the total volume of the solution to `"250 cm"^3`.