Notifications
Clear all
0
02/01/2022 5:38 pm
Topic starter
The concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an 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-1?
This topic was modified 3 years ago by admin
1 Answer
0
02/01/2022 5:45 pm
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 = \(\frac{68}{63}\)mol
= 1.079 mol
Given,
Density of solution = 1.504 g mL-1
∴ Volume of 100 g solution = \(\frac{100}{1.504}\)mL
= 66.49 mL
= 66.49 x 10-3L
Molarity of solution = \(\frac{1.079\;mol}{66.49 \times 10^{-3}L}\)
= 16.23 M