How much concentrated 5.24M sulfuric acid is need to prepare 4.4 mL of a 5.29M solution

V1= 4.44ml

Explanation:

C1= 5.24M V2=?, C2= 5.29, V2= 4.4ml

Applying dilution formula

C1V1=C2V2

5.24×V1 = 5.29×4.4

V1 =4.44ml

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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%