Concentrated nitric acid used for laboratory works 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 solution is 1.504 g mL- 1 ?

Question

Concentrated nitric acid used for laboratory works 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 solution is 1.504 g mL- 1 ? (A) 26.23M (B) 16.23M (C) 6.23M (D) 46.23M

Tardigrade Admin · Accepted Answer

68% mass of H N O3 means 100 g solution contains 68 g H N O3 ∴ Volume of solution = wt. of solution / density =(100/1.504) = 66.49 mL Molarity (M)=(m o l e o f H N O3/v o l u m e o f s o l u t i o n i n l i t r e) =(68 × 1000/63 × 66.49)=16.23M

Q. Concentrated nitric acid used for laboratory works 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 solution is $1.504 g m L^{- 1}$ ?

$26.23 M$

$16.23 M$

$6.23 M$

$46.23 M$

Q. Concentrated nitric acid used for laboratory works 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 solution is 1.504gmL−1 ?

26.23M

16.23M

6.23M

46.23M

68% mass of HNO3​ means 100 g solution contains 68 g HNO3​ ∴ Volume of solution=wt.ofsolution/density=1.504100​=66.49mL Molarity(M)=volumeofsolutioninlitremoleofHNO3​​ =63×66.4968×1000​=16.23M

Q. Concentrated nitric acid used for laboratory works 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 solution is $1.504 g m L^{- 1}$ ?

$26.23 M$

$16.23 M$

$6.23 M$

$46.23 M$