# Question #b6a41

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

So, you're dealing with a solution of sulfuric acid, **density**, **percent concentration by mass**,

The first thing to do here is use the solution's **density** to find the *mass* of the

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 **for every**

#275 color(red)(cancel(color(black)("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"color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.

Now, the problem provides you with the solution's *percent concentration by mass*, *grams of solute* you get **per**

In your case, a **for every**

#435 color(red)(cancel(color(black)("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"_4color(white)(a/a)|)))#

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

thenumber of molesof solute present in the samplethe volume of the sampleexpressed in liters

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(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

#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(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

In essence, you must find the volume of this solution that must be **diluted** to a final concentration of

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

Rearrange this equation to solve for

#c_1V_1 = c_2V_2 implies V_1 = c_2/c_1 * 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 *enough water* to make the total volume equal to