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
Explanation:
Start by calculating the number of moles of ammonia present in the
As you know, a
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
#"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
As you know, the molarity of a solution tells you the number of moles of solute present for every
In your case, the number of moles of ammonia present in
#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
#color(darkgreen)(ul(color(black)("molarity = 0.080 mol L"^(-1))))#
The answer is rounded to two sig figs.