The student repeated the investigation. The student used solutions that had different concentrations from the first investigation. The student found that 15.5 cm3 of 0.500 mol/dm3 dilute sulfuric acid completely reacted with 25.0 cm3 of potassium hydroxide solution. The equation for the reaction is: – 6105

Q1.

The student repeated the investigation. 

The student used solutions that had different concentrations from the first investigation. 

The student found that 15.5 cm3 of 0.500 mol/dm3 dilute sulfuric acid completely reacted with 25.0 cm3 of potassium hydroxide solution. 

The equation for the reaction is:

Calculate the concentration of the potassium hydroxide solution in mol/dm3 and in g/dm3 Relative atomic masses (Ar): H = 1 O = 16 K = 39

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Concentration in mol/dm3 = _____________________ mol/dm3 

Concentration in g/dm3 = ______________________ g/dm3 (6)

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One thought on “The student repeated the investigation. The student used solutions that had different concentrations from the first investigation. The student found that 15.5 cm3 of 0.500 mol/dm3 dilute sulfuric acid completely reacted with 25.0 cm3 of potassium hydroxide solution. The equation for the reaction is: – 6105

  1. Moles of H₂SO₄ =

    0.500× 15.5/1000
    ​ =0.00775 mol

    From the equation, 2 moles of KOH react with 1 mole of H₂SO₄:
    Moles of KOH=2×0.00775=0.0155 mol
    Concentration of KOH =
    moles/
    volume in dm³=
    0.0155/0.025=0.62 mol/dm³

    Relative formula mass (Mr) of KOH = 56

    Concentration in g/dm³
    =Mr×concentration in mol/dm³=56×0.62=34.7 g/dm³

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