Question #f05f4

1 Answer
Jun 23, 2016

Answer:

Here's what I got.

Explanation:

All you have to do here is do some basic molarity calculations using number of moles, #n#, and volume of solution, #V#.

Molarity, #c#, is defined as moles of solute per liter of solution. This essentially means that in order to calculate a solution's molarity, you need to figure out how many moles of solute you have in #"1 L"# of solution.

I will skip the first example because the value given to you for the number of moles is clearly inaccurate. You simply cannot have a molarity that high.

For the second example, you have a solution that has a molarity of #"2.09 mol L"^(-1)#, and are interested in figuring out how many moles of solute you get in #"4.025 L"# of solution.

In such cases, you can use the molarity of the solution as a conversion factor to help you convert the volume of the solution to number of moles of solute

#4.025 color(red)(cancel(color(black)("L solution"))) * overbrace("2.09 moles solute"/(1color(red)(cancel(color(black)("L solution")))))^(color(purple)(" = 2.09 mol L"^(-1))) = color(green)(|bar(ul(color(white)(a/a)color(black)("8.41 moles solute")color(white)(a/a)|)))#

The answer must be rounded to three sig figs you have for the molarity of the solution.

Finally, for the third example, you have #446.1# moles of solute in a volume of #"29.65 L"# of solution, which means that the molarity of the solution will be

#c = "446.1 moles"/"29.65 L" = color(green)(|bar(ul(color(white)(a/a)color(black)("15.05 mol L"^(-1))color(white)(a/a)|)))#

This time, the answer must be rounded to four sig figs.