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 1.504 g mL-'?​

Answer 1

Explanation:

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

$\sf\green{ Mass~percentage}$ (%w.w) = 68 % (w/w)

$\sf\green{ Density~of~solution}$ = 1.504 g.ml$^{-1}$

$\sf\green{ Molarity }$ (M) = $\sf\red{ ? }$

• M = $\sf\frac{ \: percentage(w/w) \: × \: 10 \: × \: d }{ GMW }$

• M = $\sf\frac{ 68~×~10~×~1.504}{ 63 }$ = 16.23 M