# 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?

Nov 22, 2016

Approx. $0.237 \cdot L$ 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,

$5.0 \cdot \cancel{L} \times 0.70 \cdot m o l \cdot \cancel{{L}^{-} 1} = 3.50 \cdot m o l$,

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

$\text{Concentration of starting solution} = 14.8 \cdot m o l \cdot {L}^{-} 1$, so we divide (i) by this concentration to get the required $\text{volume}$:

$\frac{3.50 \cdot \cancel{m o l}}{14.8 \cdot \cancel{m o l} \cdot {L}^{-} 1} = 0.237 \cdot L$. (Why are these units in $L$? Well, the $\text{moles}$ cancel out, and because $\frac{1}{1 \cdot {L}^{-} 1} = \frac{1}{\frac{1}{L}} = 1 \cdot L$).

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

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$

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.