Determine the molarity of a 6.0 mole% sulfuric acid solution with sg-a 1.07 note: atomic weight: s (32), o 16); h (o)

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

= density of water = 1 g/mL

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume =

1 mL = 0.001 L

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

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