Question

find the molarity and normality of a solution containing 4.9 grams of H2So4 present in 500 ml of solution​

Answer 1

Given:-

→ Mass of H2SO4 = 4.9g

→ Volume of solution = 500 mL

To find:-

→ Molarity of the solution.

→ Normality of the solution.

Solution:-

• Molar mass of Hydrogen (H) = 1 g/mol

• Molar mass of Sulphur (S) = 32 g/mol

• Molar mass of Oxygen (O) = 16 g/mol

Hence, molar mass of H₂SO₄ :-

= 1×2 + 32 + 16×4

= 2 + 32 + 64

= 98 g/mol

Number of mole in 4.9g of H₂SO₄ :-

= Given Mass/Molar mass

= 4.9/98

= 0.05 mole

Now let's convert the unit of volume of solution.

=> 1 mL = 0.001 L

=> 500 mL = 500(0.001)

=> 0.5 L

Molarity of a solution :-

= Moles of solute/Liters of solution

= 0.05/0.5

= 0.1 M

Thus, molarity is 0.1M .

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∵ Basicity of H₂SO₄ = 2

∴ Equivalent mass of H₂SO₄ = 98/2 = 49g

Normailty of a solution :-

= Molarity × Mol.mass/Eq. mass

= 0.1×98/49

= 0.1×2

= 0.2 N

Thus, normality is 0.2N .

Answer 2

Answer:

Molarity of the solution will be 0.1 M

Normality of the solution will be 0.2N