# 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. $0.237 \cdot L$ 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.

If you are unsatisfied, voice your obejctions, and someone here will help you.