Question

Question #b6a41

Answer 1

Here's what I got.

Explanation:

So, you're dealing with a solution of sulfuric acid, `"H"_2"SO"_4`, of known density, `"1.58 g mL"^(_1)`, and percent concentration by mass, `"68.50% m/m"`.

The first thing to do here is use the solution's density to find the mass of the `"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 `"1.58 g mL"^(-1)` tells you that you get `"1.58 g"` for every `"1 mL"` of solution. This means that your sample will have a mass of

`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, `"% m/m"`, which essentially tells you how many grams of solute you get per `"100 g"` of solution.

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

`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

• 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

`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 `"1.5 L"` of a `"3.5 M"` solution.

In essence, you must find the volume of this solution that must be diluted to a final concentration of `"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_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 `"480 mL"` of the `"11.0 M"` sulfuric acid solution and add enough water to make the total volume equal to `"1.5 L"`, you'll get a `"3.5 M"` sulfuric acid solution.