Question

What volume of `84%` `HNO_3` `"w/w"`, `rho=1.54*g*mL^-1`, is required to make a `1.0*L` of a `1*mol*L^-1` solution of nitric acid?

Answer 1

Approx. `25*mL`

Explanation:

We use the relationship, `"concentration"="moles of solute"/"volume of solution"`

And thus we need `1*Lxx0.50*mol*L^-1=0.50*mol` with respect to nitric acid. And this represents a mass of `0.50*molxx63.01*g*mol^-1=31.5*g`.

We gots `84%` `HNO_3`, `"w/w"`............

And thus we need a mass of solution......

`(31.5*g)/(0.84)=37.5*g`.......

And given the density......`rho=1.54*g*mL^-1`, we need....

`(37.5*g)/(1.54*g*mL^-1)=24.4*mL.`

Nitric acid of this concentration is called in the trade `"red fuming nitric acid"`, and its density owes to the solubility of `NO_2`.