What was the concentration of my stock solution if I used 23mL of it to make 102 ml of 1.4 M dilution?

formula:c1v1=c2v2

Explanation:

v1=102ml,c1=1.4mol/l

v2=?;c2=2.4mol/l

v2=v1×c1÷c2=102ml×1.4mol/l÷2.4mol/l

=59.5ml

i hope this will help you

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%