Question

A 30.00% (by mass) solution of `HNO_3` in water has a density of 1.18 g/cm^3 at 20°C. What is the molarity of `HNO_3` in the solution?

Answer 1

We are given that `"Mass of nitric acid"/"Mass of solution"xx100%` `=` `30.00%`.

Explanation:

Now `"Concentration"="Moles of solute"/"Volume of solution"`.

How do we turn one t'other? Well, we need the density of the solution, which you have supplied, `rho=1.18*g*cm^-3`, and some tedious 'rithmetic................

So we work with a `1*L` volume of solution, i.e. `10^3*cm^3`.

`"Moles of solute"=(30.00%xx10^3*cancel(cm^3)xx1.180*cancelg*cancel(cm^-3))/(63.01*cancelg*mol^-1)`

`=5.62*mol`

But this molar quantity was dissolved in a `1L` volume of solvent.....

`"Concentration"=(5.62*mol)/(1*L)=5.62*mol*L^-1`.

Now I happen to know that conc. nitric acid, which is commercially supplied as a `68%(w/w)` solution, has a molar concentration of approx. `15*mol*L^-1`, so the value we calculated is reasonably consistent. You should keep doing these types of problems; it's all too easy to get bewildered and make an error.