Question

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?

Answer 1

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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