Question #1f4c5

1 Answer
Oct 18, 2017

Answer:

#"4.00 mol L"^(-1)#

Explanation:

For starters, you know that the density of pure water is approximately equal to #"1.00 g mL"^(-1)#. By comparison, this aqueous solution has a density of #"1.25 g mL"^(-1)#.

The idea here is that the difference between the mass of #"1 mL"# of pure water and the mass of #"1 mL"# of this solution will represent the mass of the solute.

For #"1 mL"# of this solution, you have

#overbrace("1.25 g")^(color(blue)("mass of solution")) - overbrace("1.00 g")^(color(blue)("mass of solvent")) = overbrace("0.25 g")^(color(blue)("mass of solvent"))#

Use the molar mass of the solvent to find the number of moles present in #"1 mL"# of this solution.

#0.25 color(red)(cancel(color(black)("g"))) * "1 mole solute"/(62.5color(red)(cancel(color(black)("g")))) = "0.00400 moles solute"#

As you know, the molarity of the solution tells you the number of moles of solute present in #10^3color(white)(.)"mL" = "1 L"# of the solution.

Since you know that #"1 mL"# contains #0.00400# moles of solute, you can say that #10^3# #"mL"# will contain

#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.00400 moles solute"/(1color(red)(cancel(color(black)("mL solution")))) = "4.00 moles solute"#

You can thus say that the molarity of the solution is equal to

#color(darkgreen)(ul(color(black)("molarity = 4.00 mol L"^(-1))))#

The answer is rounded to three sig figs.