The concentration of a solution of ammonia, #"NH"_3# is #"1.5% m/v"#. What is the molar concentration of a solution produced by diluting #"25.0 mL"# of this solution with #"250 mL"# of water?

1 Answer
Jul 27, 2017

Answer:

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

Explanation:

Start by calculating the number of moles of ammonia present in the #"25.0-mL"# sample of #"1.5% m/v"# ammonia solution.

As you know, a #"1.5% m/v"# solution contains #"1.5 g"# of solute, which in your case is ammonia, for every #"100 mL"# of solution.

This means that your sample will contain

#25.0 color(red)(cancel(color(black)("mL solution"))) * "1.5 g NH"_3/(100color(red)(cancel(color(black)("mL solution")))) = "0.375 g NH"_3#

Use the molar mass of ammonia to convert this to moles

#0.375 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17.031color(red)(cancel(color(black)("g")))) = "0.02202 moles NH"_3#

Now, you're diluting this sample by adding #"250 mL"# of water. The total volume of the diluted solution will be

#"25.0 mL + 250 mL = 275 mL"#

Since you only added water, the number of moles of ammonia will remain unchanged, i.e. the diluted solution contains the same number of moles as the #"25.0-mL"# sample.

As you know, the molarity of a solution tells you the number of moles of solute present for every #"1 L" = 10^3# #"mL"# of solution.

In your case, the number of moles of ammonia present in #10^3# #"mL"# of this diluted solution will be equal to

#10^3 color(red)(cancel(color(black)("mL solution"))) * overbrace("0.02202 moles H"_3/(275 color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the known composition of the diluted solution")) = "0.080 moles NH"_3#

So, if you have #0.080# moles of ammonia in #10^3color(white)(.)"mL" = "1 L"# of the diluted solution, you can say that the molarity of the solution is equal to

#color(darkgreen)(ul(color(black)("molarity = 0.080 mol L"^(-1))))#

The answer is rounded to two sig figs.