Question

Nitric acid is usually purchased in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g/mL. How much concentrated solution would you take to prepare 1.00 L of 0.115 M HNO3 by mixing with water?

Answer 1

Let volume of solution be 1 L
volume , V = 1 L
= 1*10^3 mL


density, d = 1.41 g/mL
use:
mass = density * volume
= 1.41 g/mL *1*10^3 mL
= 1.41*10^3 g
This is mass of solution
mass of HNO3 = 70.3 % of mass of solution
= 70.3*1410.0/100
= 991.23 g

Molar mass of HNO3,
MM = 1*MM(H) + 1*MM(N) + 3*MM(O)
= 1*1.008 + 1*14.01 + 3*16.0
= 63.018 g/mol


mass(HNO3)= 991.23 g

use:
number of mol of HNO3,
n = mass of HNO3/molar mass of HNO3
=(9.912*10^2 g)/(63.02 g/mol)
= 15.73 mol
volume , V = 1 L


use:
Molarity,
M = number of mol / volume in L
= 15.73/1
= 15.73 M

use dilution formula
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution

Given:
M1 = 15.73 M
M2 = 0.115 M
V2 = 1.00 L = 1000 mL

use:
M1*V1 = M2*V2
V1 = (M2 * V2) / M1
V1 = (0.115*1000)/15.73
V1 = 7.31 mL
Answer: 7.31 mL