You wish to make a 0.405 M hydrochloric acid solution from a stock solution of 3.00 M hydrochloric acid. How much concentrated acid must you add to obtain a total volume of 100 mL of the dilute solution?

Explanation:

We have to make 100 mL 0f .405M HCl from 3 M solution of HCl .

volume of 3M to be taken required . Let this volume be V litre .

V litre of 3M will contain 3 V moles of HCl .

moles contained by 100 mL of ,405 HCl = .405 x .1 = .0405 moles .

So 3V = .0405

V = .0405 / 3 litre

= .0405 x 1000 / 4 mL

= 10.125 mL

So we have to take 10.125 mL of 3M HCl and add water to them to make its volume 100 mL .

helloi think dog cat bunny dog cat bunny dog cat bunny

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%