# What volume of concentrated HCl contains a mass of 125*g HCl?

Sep 20, 2016

34% $\frac{w}{w}$ $H C l$ is $10.9 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

The conc. acid you use in a lab is 34% $\frac{w}{w}$. As a I recall, this has a density of approx. $1.17 \cdot g \cdot m {L}^{-} 1$.

We require $125 \cdot g \cdot H C l$ $=$ $\frac{125 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.43 \cdot m o l$.

For volume, we divide the molar quantity by the concentration:

$\frac{3.43 \cdot \cancel{m o l}}{10.9 \cdot \cancel{m o l} \cdot \cancel{{L}^{-} 1}} \times 1000 \cdot m L \cdot \cancel{{L}^{-} 1}$ $=$ $315 \cdot m L$

But of course the questioner has not specified the concentration with which we work. We are guessing.

Working back, a volume of $283 \cdot m L$ corresponds to $12.2 \cdot m o l \cdot {L}^{-} 1$. This is very concentrated muriatic acid. The concentration of the acid should have been specified as a boundary condition.

Sep 20, 2016

It should be given here that w/w%of conc. Hydrochloric acid is 37.2%
The calculation of volume will be as fillows.

As the density of 37.2%(w/w) is $1.19 \frac{g}{m l}$ the volume of 100g 37.2% conc hydrochloric acid will be $\frac{100}{1.19} \left(m l\right)$ and this volume of conc. acid will contain $37.2 g H C l$

Hence $125 g$ of conc.hydrochloric acid will be available from the volume
$\frac{100}{1.19} \times \frac{125}{37.2} m l \approx 282.37 m l \text{ HCl}$

Sep 20, 2016

It depends what figure you take as the molarity of concentrated hydrochloric acid. I took it as 12.19 M which gives an answer of 281.26 ml (which is roughly 282 ml!).

#### Explanation:

Concentrated hydrochloric acid has concentration of 12.19 mol/litre,

The molar mass of HCl is 36.458 g/mol.

125 g is therefore (125/36.458) = 3.4286 moles.

So the volume that would contain this number of moles is going to be (3.4286 / 12.19) = 0.28126 litres or 281.26 ml.