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 1.30 L of 0.105 M HNO3by mixing with water?

Answer 1

en 14 I 30 . 30X 0-105 0.1365 mo) 3 0.1345 X 63.0) = 8.600 to-3 2. 23 2 23 j.4丿 8.67

Answer 2

First, find the number of moles of nitric acid in the diluted solution by multiplying the volume of the solution by its molarity; this amount is equal to the number of moles of nitric acid in the concentrated solution:

Use the molar mass of nitric acid as a conversion factor between the number of moles and the mass of nitric acid. Then, use the mass percent concentration of the concentrated solution as a conversion factor between the mass of nitric acid and the mass of this solution. Finally, use the density of the concentrated solution as a conversion factor between the mass of this solution and its volume. Therefore,

Answer 3

First, find the number of moles of nitric acid in the diluted solution by multiplying the volume of the solution by its molarity; this amount is equal to the number of moles of nitric acid in the concentrated solution:

HNO3=1.30L×0.120 mol/ 1 L=0.156 mol

Use the molar mass of nitric acid as a conversion factor between the number of moles and the mass of nitric acid. Then, use the mass percent concentration of the concentrated solution as a conversion factor between the mass of nitric acid and the mass of this solution. Finally, use the density of the concentrated solution as a conversion factor between the mass of this solution and its volume. Therefore,

0.156mol * 63g/mol * 100g solution/70.3g * 1mL/1.41g solution = 9.92