How do we make a #5.0*L# volume of aqueous ammonia of #0.70*mol*L^-1# concentration, from conc. ammonia of #14.8*mol*L^-1# concentration?

1 Answer
Nov 22, 2016

Approx. #0.237*L# conc. ammonia are required.

Explanation:

Ammoniacal solutions tend to have a really pungent smell; the kind that brings tears to your eyes when you get a whiff of them. Ammonia is present in many household cleaning products.

We work out (i), the moles of ammonia present in the required solution,

#5.0*cancel(L)xx0.70*mol*cancel(L^-1)=3.50*mol#,

And (ii) the volume of conc. ammonia that contains that molar quantity:

#"Concentration of starting solution"=14.8*mol*L^-1#, so we divide (i) by this concentration to get the required #"volume"#:

#(3.50*cancel(mol))/(14.8*cancel(mol)*L^-1)=0.237*L#. (Why are these units in #L#? Well, the #"moles"# cancel out, and because #1/(1*L^-1)=1/(1/L)=1*L#).

The problem spoon-feeds you a bit, but you are required to know the definition of concentration:

#"Concentration"# #=# #"Moles of solute"/"Volume of solution"#

I have been careful to do the calculation dimensionally thruout. I wanted an answer in #L#, the calculation gave me an answer in #L#. It's all too easy to divide when we should have multiplied, or vice versa, and the inclusion of units in our calculations is an extra check on our approach.

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