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 1.504gml-1

Answer 1

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution.

Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol - 1

Then, number of moles of HNO3 = 68 / 63 mol

= 1.08 mol

Also density = 1.504g/mL-1 (given)

Therefore from the formula density = mass / volume,

we get

Volume of solution = 1000/1.504 = 66.49 mL

Therefore molarity of nitric acid = (1.08/66.49) x 1000 = 16.24 M

Answer 2

$\fbox{\texttt{\green{Answer:}}}$

Molarity of the solution(sample) = 16.23 M

$\fbox{\texttt{\pink{Given:}}}$

68% nitric acid means that :-

Mass of nitric acid = 68 g

Mass of solution = 100 g

Density of the solution = $\bold{1.504\:g\:mL^{-1}}$

$\fbox{\texttt{\blue{To\:find:}}}$

Molarity of the solution (sample).

$\fbox{\texttt{\red{How\:to\:Find:}}}$

Molar mass of $\bold{HNO_3\:=\:63\:g\:mol^{-1}}$

$\therefore\:\bold{68\:g\:HNO_3=\frac{68}{63}\:mole=1.079\:mole}$

Density of the solution = $\bold{1.504\:g\:mL^{-1}}$ (given)

$\therefore\:$Volume of solution =

$\bold{\frac{100}{1.504}\:mL=66.5\:mL=0.0665\:L}$

$\bold{Molarity\:of\:the\:solution =\frac{Moles\:of\:the\:solute}{Volume\:of\:solution\:in\:L}\:=}$

$\bold{\frac{1.079}{0.0665}\:M=16.23\:M}$