Question

(i) Concentrated nitric acid used as a laboratory reagent is usually 69% by mass of nitric acid. Calculate the volume of the solution which contained 23g of HNO3. Density of conc. HNO3 solution is 1.41g cm^-3

Answer 1

Mass percent of nitric acid in the sample = 69 % [Given]


Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.


Molar mass of nitric acid (HNO3)


= {1 + 14 + 3(16)} g mol–1


= 1 + 14 + 48


= 63 g mol-¹
=

 


∴Concentration of nitric acid = 15.44 mol/L



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

The concentration of the solution is 15.44 M.


 

Answer 2

Answer:The volume of the solution which contained 23 grams of nitric acid is 23.47 mL.

Explanation:

Given that concentrated nitric acid used as a laboratory reagent is usually 69% by mass of nitric acid.

Mass of nitric acid = 23 g

Mass of the solution = M

$69\%=\frac{23 g}{M}\times 100$

M = 33.33 g

Density of the solution = $1.41 g/cm^3=1.42 g/ml$

$Density=\frac{Mass}{Volume}$

$V=\frac{M}{1.42 g/mL}=\frac{33.33}{1.42 g/mL}$

V = 23.47 mL

The volume of the solution which contained 23 grams of nitric acid is 23.47 mL.