Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41g/ml and the mass percent of nitric acid in it being 69%.
Answers
Answer:
given: 1.41 g/mL HNO3 solution
the solution is 69% HNO3
1. calculate the mass of HNO3 in the solution
= 1.41 g/ml x 69%
= 0.9729 g/ml
2. convert the answer in 1 to mole quantity by dividing it with the molar mass of HNO3 which is 63 g/mol
= 0.9729 g/ml divided 63 g/mol
= 0.01544 mol/ml
3. Since molar quantity is expressed as mole per L, you need to multiply the answer in 2 with 1000 (1L = 1000ml)
= 0.01544 mol/ml X 1000
= 15.44 M
Answer: 15.44 M
Explanation: Given density of aqueous solution is 1.41 gm/ml of which 69% is nitric acid (HNO3).
So, 69% of 1.41 => 1.41 x 69/100 = 0.9729
Density of Nitric Acid = 0.9729 gm/ml
Now, Density = Mass/Volume. So,
0.9729 = Mass of HNO3/1000ml
Mass of HNO3 = 972.9 gm
No. of moles of HNO3 = Mass of HNO3/Molar mass of HNO3
= 972.9/63
= 15.44 mol.
So Molarity of HNO3 = No. of moles of HNO3/Volume of Solution
= 15.44/1 L
= 15.44 M
(Note) : We find molarity because in the question it is asked to moles per litre which is molarity(M).