Question

You can purchase nitric acid in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g/mL. Describe exactly how you would prepare 1.15 L of 0.100 M HNO3 from the concentrated solution.

Answer 1

Answer:

You will need to dilute 7.30 ml of the concentrated solution into the 1.15 L of the diuluted one by adding water.

Explanation:

First you need to calculate how much acid you'll need in order to prepare the diluted solution. If you know the volume and the molarity, just multiply.

1.15 l * 0.1 M = 0.115 mols of HNO3

Multiplying by the molar mass you'll convert that value into grams

0.115 * 63g/mol = 7.245 g HNO3

Now you use the concentrated solution to know how much to extract.

70.3 g of HNO3 pure are in 100 g of Solution, then 7.245 are in x

X= (7.245 * 100)/ 70.3 = 10.3 grams of concentrated solution.

Using the density you convert the mass into volume:

V = m/dens --> V = 10.3 g / 1.41 g/ml --> V= 7.3 ml

You'll take 7,30 ml of the concentrated solution and add water until reach the required volume for the solution.

Answer 2

The way to working out the numbers is to increase the measure of HNO3 required by the molarity to discover what number of moles you require: 0.115. You ought to have the capacity to make sense of the recipe weight H is 1, N is 14, O is 16. The result of the quantity of moles duplicated by the recipe weight ought to give an esteem in grams. You can utilize the thickness to change over to a volume of HNO3 to add to the right volume of water.