Question #b6a41

1 Answer
Apr 16, 2016

Answer:

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.