In acid dilution problem, describe how you created 500 ml of a 5.5 M HCl solution from the 11.6 M stock solution. Show your calculations and give details about the equipment used, which solutions you mixed together, and amounts of each.

237 ml of stock solution is taken in a 500 ml volumetric flask and diluted with water until the 500 ml mark to obtain 5.5 M diluted solution

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

where,

= molarity of stock solution = 11.6 M

= volume of stock solution = ?

= molarity of dilute solution = 5.5 M

= volume of dilute solution = 500 ml

Thus 237 ml of stock solution is taken in a 500 ml volumetric flask and diluted with water until the 500 ml mark to obtain 5.5 M diluted solution

chemistry sucks a**

explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%