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ou are required to prepare 500 mL of a solution containing 0.02 molar Sulphuric Acid. How much Dilute Sulphuric Acid would you use?Note:Dilute Sulphuric Acid contains 10% w/w Sulphuric acid; density = 1.045 g/mlH2SO4; MW = 98.07

Question

ou are required to prepare 500 mL of a solution containing 0.02 molar Sulphuric Acid. How much Dilute Sulphuric Acid would you use?Note:Dilute Sulphuric Acid contains 10% w/w Sulphuric acid; density = 1.045 g/mlH2SO4; MW = 98.07

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Solution

To prepare a 0.02 M solution of sulphuric acid in 500 mL, you need to know the amount of sulphuric acid required.

  1. First, calculate the moles of sulphuric acid needed. Molarity (M) = moles/volume(L), so moles = Molarity * volume(L) = 0.02 M * 0.5 L = 0.01 moles.

  2. Convert moles to grams using the molecular weight of sulphuric acid (H2SO4), which is 98.07 g/mole. So, grams = moles * molecular weight = 0.01 moles * 98.07 g/mole = 0.9807 g.

  3. Now, you need to find out how much of the 10% w/w sulphuric acid solution contains this amount of sulphuric acid. In a 10% w/w solution, 100 g of solution contains 10 g of sulphuric acid. So, 0.9807 g of sulphuric acid is in (0.9807 g / 10 g) * 100 g = 9.807 g of the solution.

  4. Convert the weight of the solution to volume using the given density (1.045 g/mL). So, volume = weight / density = 9.807 g / 1.045 g/mL = 9.38 mL.

So, you would use approximately 9.38 mL of the 10% w/w sulphuric acid solution to prepare 500 mL of a 0.02 M sulphuric acid solution.

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