Question a14ed

Mar 27, 2016

${\text{8.43 mol L}}^{- 1}$

Explanation:

The most important thing to remember when dealing with a dilution is that the number of moles of solute, i.e. the amount of solute, you have in the initial solution remains constant after the dilution is performed.

In essence, diluting a solution implies increasing its volume while keeping the number of moles of solute constant.

Since molarity is defined as the number of moles of solute per liter of solution, increasing the volume while keeping the number of moles of solute constant will result in a decrease in concentration.

color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liters of solution"color(white)(a/)|)))

In this case, you know that the diluted solution has a molarity of ${\text{2.68 moles L}}^{- 1}$, which implies that every liter of solution contains $2.68$ moles of solute, which in your case is sulfuric acid.

So, you know that this solution contains $2.68$ moles of sulfuric acid.

Since you performed a dilution, you can say for a fact that this is how many moles of sulfuric must have been present in the initial solution.

This means that the initial solution had a molarity of

c_"initial" = "2.68 moles"/(318 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"8.43 mol L"^(-1)color(white)(a/a)|)))

The answer is rounded to three sig figs.

This is exactly what you get by using the equation for dilution calculations, which looks like this

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

Here you have

${c}_{1}$, ${V}_{1}$ - the molarity and volume of the concentrated solution
${c}_{2}$, ${V}_{2}$ - the molarity and volume of the diluted solution

If you rearrange the equation to solve for ${c}_{1}$, you will get

${c}_{1} {V}_{1} = {c}_{2} {V}_{2} \implies {c}_{1} = {V}_{2} / {V}_{1} \cdot {c}_{2}$

c_1 = (1.00 color(red)(cancel(color(black)("L"))))/(318 * 10^(-3)color(red)(cancel(color(black)("L")))) * "2.68 M" = color(green)(|bar(ul(color(white)(a/a)"8.43 mol L"^(-1)color(white)(a/a)|)))#

It's worth noting that the ratio between the final volume and the initial volume of the solution is called the dilution factor.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{D.F." = V_"final"/V_"initial} \textcolor{w h i t e}{\frac{a}{a}} |}}}$