Question

Density of 15% (by weight)H2SO4 solution is 1.5g/ml then molarity of solution is:-

Answer 1

Answer:

amount of solute is 15% ( w/v)

∴ 15g of solute (H₂SO₄) is present in 100 mL of solution

But density of solution is 1.1 g/cm³

Hence, mass of solution = volume of solution × density of solution

= 100mL × 1.1 g/mL [ ∵ 1cm³ = 1 mL ]

= 110g

∴ mass of solvent = mass of solution - mass of solute

= 110g - 15g = 95g

Now, molality = mole of solute × 1000/mass of solvent in g

= {weight of solute} × 1000/molecular mass of solute × mass of solvent

= 15 × 1000/98 × 95 [ ∵ molecular mass of H₂SO₄ = 98 g/mol

= 1.61

Hence, molality = 1.61

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Explanation:

Answer 2

Given:

Weight Percent of solution, wt % = 15 %

The density of solution, d = 1.5 gm / ml

To Find:

The molarity of the given solution.

Calculation:

- For H2SO4, M. wt. = 98 gm / mol

- We know that molarity can also be given as:

M = (10 × d × wt %) / M. wt

⇒ M = (10 × 1.5 × 15) / 98

⇒ M = 225 / 98

⇒ M = 2.296 ≈ 2.3 M

- So, the molarity of the given solution is 2.3 M.