Question

Concentrated nitric acid used in laboratory work is $68\%$ nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is $1.504gm{L^{ - 1}}$?

Answer 1

Hint – Start by describing the general concepts behind moles, density and molarity. - Then assume that you start off with $100g$ of the given sample and calculate the amount of nitric acid ($68\%$ of $100g$). Then use the equation $No{\text{ }}of{\text{ }}moles = \dfrac{{Weight}}{{Molar\;mass}}$ to find the number of moles of nitric acid. Then use the equation for density to find out the volume i.e.$Density = \dfrac{{Weight}}{{Volume}}$.
- Then use $Molarity = \dfrac{{Number{\text{ }}of{\text{ }}moles}}{{Volume}}$ to find out the molarity.

Complete step step solution:

Before attempting the solution, we should describe what mole, density and molarity is.
So, 1 mole is the amount of a substance that has exactly $6.02214076 \times {10^{23}}$atoms or particles of that substance.
- Density ($d$) – It is the measurement of the amount (in$kg$) of a substance in unit volume (in$L$) of the solution. The SI unit of density is$kg/L$.
- Molarity ($M$) – It is a measurement of concentration of a certain substance in a solution.
- Molarity is by definition the moles of a solute that is present in a liter of the solution.
- Molarity is called the molar concentration of a solution. The unit of molarity is $moles/L$.
Let the weight of the solution be$100g$. So given in the problem is that$100g$ of the solution contains $68\%$nitric acid by mass in aqueous solution i.e. $68g$.
We know that
$No{\text{ }}of{\text{ }}moles = \dfrac{{Weight}}{{Molar\;mass}}$
No of moles of nitric acid $= \dfrac{{68}}{{63}}$($\therefore$Molar mass of nitric acid is 63)
$\Rightarrow n = 1.079{\text{ }}moles{\text{ }}of{\text{ }}nitric{\text{ }}acid$
Now, we know
$Density = \dfrac{{Weight}}{{Volume}}$
$\Rightarrow 1.504 = \dfrac{{100}}{V}$
$\Rightarrow V = \dfrac{{100}}{{1.504}}$
$\Rightarrow V = 66.5mL\;or0.0665L$
We know that the equation for molarity is
$Molarity = \dfrac{{Number{\text{ }}of{\text{ }}moles}}{{Volume}}$
$\Rightarrow M = \dfrac{{1.079}}{{0.0665}}$
$\Rightarrow M = 16.22moles/liter$
Hence, the molarity of the given sample is $16.22moles/liter$

Note – 68% Nitric acid solution in water is an azeotropic mixture , which is a constant boiling mixture in which concentration of solute and solvent remains constant in solution as well as in vapour phase.