Question

75 ml of H2SO4 (specific gravity=1.18) containing 49% H2SO4 by mass is diluted to 590 ml.Calculate the molarity of diluted solution.

Answer 1

Hello Dear.

Given Conditions ⇒
Volume of the Sulphuric Acid solution = 75 ml.
Specific Gravity = 1.18

∵ Specific gravity is the ratio of the Density of the Substance to the Density of the water.
∴ 1.18 = Density of the Sulphuric solution/1 g/ml.
∴ Density of the Sulphuric Acid solution= 1.18 g/ml.

Now, Using the Formula,
Density = Mass/Volume.
∴ Mass = Density × Volume.
∴ Mass = 1.18 × 75
⇒ Mass = 88.5 g.

Mass of the sulphuric acid = 49/100 × 88.5
= 43.365 g.

We know, Molar mass of the Sulphuric Acid = 98 g/mol.

∴ No. of moles of H₂SO₄ = Mass/Molar Mass.
∴ No. of moles = 43.365/98
⇒ No. of moles = 0.4425 moles.

Volume of the solution = 590 ml.
= 0.590 liter.

Now, Using the Formula,
 Molarity = No. of moles of Sulphuric Acid/Volume of the solution
∴ Molarity = 0.4425/0.590
⇒ Molarity = 0.75 M.


Hence, the Molarity of the solution is 0.75 M.


Hope it helps.

Answer 2

Answer:

Given Conditions ⇒

Volume of the Sulphuric Acid solution = 75 ml.

Specific Gravity = 1.18

∵ Specific gravity is the ratio of the Density of the Substance to the Density of the water.

∴ 1.18 = Density of the Sulphuric solution/1 g/ml.

∴ Density of the Sulphuric Acid solution= 1.18 g/ml.

Now, Using the Formula,

Density = Mass/Volume.

∴ Mass = Density × Volume.

∴ Mass = 1.18 × 75

⇒ Mass = 88.5 g.

Mass of the sulphuric acid = 49/100 × 88.5

= 43.365 g.

We know, Molar mass of the Sulphuric Acid = 98 g/mol.

∴ No. of moles of H₂SO₄ = Mass/Molar Mass.

∴ No. of moles = 43.365/98

⇒ No. of moles = 0.4425 moles.

Volume of the solution = 590 ml.

= 0.590 liter.

Now, Using the Formula,

 Molarity = No. of moles of Sulphuric Acid/Volume of the solution

∴ Molarity = 0.4425/0.590

⇒ Molarity = 0.75 M.

Hence, the Molarity of the solution is 0.75 M