Question

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

Answer 1

`34%` `w/w` `HCl` is `10.9*mol*L^-1`

Explanation:

The conc. acid you use in a lab is `34%` `w/w`. As a I recall, this has a density of approx. `1.17*g*mL^-1`.

We require `125*g*HCl` `=` `(125*g)/(36.46*g*mol^-1)` `=` `3.43*mol`.

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

`(3.43*cancel(mol))/(10.9*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)` `=` `315*mL`

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

Working back, a volume of `283*mL` corresponds to `12.2*mol*L^-1`. This is very concentrated muriatic acid. The concentration of the acid should have been specified as a boundary condition.

Answer 2

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 `1.19g/(ml)` the volume of 100g `37.2%` conc hydrochloric acid will be `100/1.19(ml)` and this volume of conc. acid will contain `37.2g HCl`

Hence `125g` of conc.hydrochloric acid will be available from the volume
`100/1.19xx125/37.2 ml~~282.37ml" HCl"`

Answer 3

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.