# Question b6a41

Apr 16, 2016

Here's what I got.

#### Explanation:

So, you're dealing with a solution of sulfuric acid, ${\text{H"_2"SO}}_{4}$, of known density, ${\text{1.58 g mL}}^{_ 1}$, and percent concentration by mass, $\text{68.50% m/m}$.

The first thing to do here is use the solution's density to find the mass of the $\text{275-mL}$ sample given to you.

The thing to remember about a substance's density is that you can use it a conversion factor to go from mass to volume and vice versa.

A density of ${\text{1.58 g mL}}^{- 1}$ tells you that you get $\text{1.58 g}$ for every $\text{1 mL}$ of solution. This means that your sample will have a mass of

$275 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL solution"))) * overbrace("1.58 g"/(1color(red)(cancel(color(black)("mL solution")))))^(color(purple)("given density")) = color(green)(|bar(ul(color(white)(a/a)"435 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

Now, the problem provides you with the solution's percent concentration by mass, $\text{% m/m}$, which essentially tells you how many grams of solute you get per $\text{100 g}$ of solution.

In your case, a $\text{68.50% m/m}$ sulfuric acid solution will contain $\text{68.50 }$ of sulfuric acid for every $\text{100 g}$ of solution. Since you know that your $\text{275 mL}$ sample has mass of $\text{435 g}$, you can say that this sample will contain

$435 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g solution"))) * overbrace(("68.50 g H"_2"SO"_4)/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("= 68.50% m/m")) = color(green)(|bar(ul(color(white)(a/a)"298 g H"_2"SO}}_{4} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In order to find this solution's molarity, you need to know two things

• the number of moles of solute present in the sample
• the volume of the sample expressed in liters

You already know the volume of the sample in milliliters, so use the known conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 L" = 10^3"mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to convert it to liters. You will have

275 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.275 L"

To find the number of moles of sulfuric acid, use its molar mass

298 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"SO"_4)/(98.08color(red)(cancel(color(black)("g")))) = "3.038 moles H"_2"SO"_4

The solution's molarity will be

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = {n}_{\text{solute"/V_"solution}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you have

c = "3.038 moles"/"0.275 L" = color(green)(|bar(ul(color(white)(a/a)"11.0 mol L"^(-1)color(white)(a/a)|)))

Finally, you need to figure out how much of this solution will be needed in order to make $\text{1.5 L}$ of a $\text{3.5 M}$ solution.

In essence, you must find the volume of this solution that must be diluted to a final concentration of $\text{3.5 M}$.

As you know, a dilution is characterized by the fact that the number of moles of solute must be kept constant. More specifically, a solution is diluted by increasing its volume while keeping the number of moles of solute constant.

Mathematically, this can be expressed as

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

Rearrange this equation to solve for ${V}_{1}$

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

Plug in your values to find

V_1 = (3.5 color(red)(cancel(color(black)("M"))))/(11.0color(red)(cancel(color(black)("M")))) * "1.5 L" = "0.4773 L"

Expressed in milliliters and rounded to two sig figs, the number of sig figs you have for the molarity and volume of the target solution, the answer will be

"volume needed" = color(green)(|bar(ul(color(white)(a/a)"480 mL"color(white)(a/a)|)))#

So, if you start with $\text{480 mL}$ of the $\text{11.0 M}$ sulfuric acid solution and add enough water to make the total volume equal to $\text{1.5 L}$, you'll get a $\text{3.5 M}$ sulfuric acid solution.