Question

Determine the percent sulfuric acid by mass of a 1.61 m aqueous solution of H2SO4. %

Answer 1

Answer:

Wt%H₂SO₄ = 10.2% (w/w)

Explanation:

1.61 molal H₂SO₄ = 1.61 mole H₂SO₄/Kg Solvent = 1.61 mole(98 g/mole)/Kg Solvent = 113.68 g H₂SO₄/Kg Solvent

Solution weight = 1000 g solvent + 113.68 g solute = 1113.68 g Solution

Wt%H₂SO₄ = (113.68 g/1113.68g)100% = 10.2% (w/w)

Answer 2

The percentage of sulfuric acid = 10.2% (w/w)

Given:

Molality of solution = 1.61m

Molality:

It is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.

1.61 molal H₂SO₄ = 1.61 mole H₂SO₄/Kg Solvent

= 1.61 mole(98 g/mole)/Kg Solvent = 113.68 g H₂SO₄/Kg Solvent

Solution is made up of solute and solvent. Thus,

Solution weight = 1000 g solvent + 113.68 g solute = 1113.68 g Solution

$Wt\% \text{ of } H_2SO_4 = \frac{113.68 g}{1113.68g}*100\% = 10.2\% (w/w)$

Find more information about Molality here:

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