Question

Question #b28e9

Answer 1

`"1.4 L"`

Explanation:

The idea here is that you need to use the molarity and volume of the target solution to determine how many moles of solute it contains.

Since you're dealing with a dilution, you know for a fact that the number of moles of solute present in the initial solution must be equal to the number of moles of solute present in the target solution.

A dilution essentially decreases the concentration of a solution by increasing the volume of the solution while keeping the number of moles of solute constant.

Calculate the number of moles of solute present in the target solution by using

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))`

In your case, you have

`n_"solute" = "0.10 mol" color(red)(cancel(color(black)("L"^(-1)))) * 3.5 color(red)(cancel(color(black)("L"))) = "0.35 moles solute"`

Since you know that this is exactly how many moles of solute must have been present in the initial solution, you can use its concentration to find the volume that would contain this many moles

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c)color(white)(a/a)|)))`

This will get you

`V_"initial" = (0.35 color(red)(cancel(color(black)("moles"))))/(0.25 color(red)(cancel(color(black)("mol"))) "L"^(-1)) = color(green)(|bar(ul(color(white)(a/a)"1.4 L"color(white)(a/a)|)))`

So, if you take `"1.4 L"` of the `"0.25-M"` solution, and add enough water to get its total volume to `"3.5 L"`, you'll get a `"0.10 M"` solution.

ALTERNATIVE APPROACH

More often than not, dilution calculations will make use of the following equation

`color(blue)(overbrace(c_1 xx V_1)^(color(green)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(green)("moles of solute in diluted solution"))`

Here

`c_1`, `V_1` - the molarity and volume of the concentrated solution
`c_2`, `V_2` - the molarity and volume of the diluted solution

This equation uses the same underlying principle of a dilution `->` the number of moles of solute must remain constant.

Rearrange to solve for `V_1`

`c_1V_1 = c_2V_2 implies V_1 = c_2/c_1 * V_2`

Plug in your values to find

`V_1 = (0.10 color(red)(cancel(color(black)("mol L"^(-1)))))/(0.25color(red)(cancel(color(black)("mol L"^(-1))))) * "3.5 L" = color(green)(|bar(ul(color(white)(a/a)"1.4 L"color(white)(a/a)|)))`