Question

Here's the other part!

(ii) Find the molar concentration of the sulfuric acid from the titration result:
Complete the calculation given below:
No. of moles of KOH in 25.00 cm3 =
= ………………………… (1 mark)
 no. of moles H2SO4 in 15.00 cm3 =
= ………………………… (1 mark)
 molar concentration of H2SO4 =
(Include units in your answer) = ………………………… (2 marks)

(Total 10 marks)

Answer 1

Relative Molecular Mass (RMM) of

KOH is 56.11 g/mole

No. of moles of KOH in 250 cm3$= \frac{0.95}{56.11}=0.0169moles$

2KOH (aq) + H₂SO₄ (aq) → K₂SO₄ (aq) + 2H₂O (l)

Moles of $H_2SO_4$ $=\frac{0.0169}{2}=0.00845moles$

Molarity =$\frac{0.00845}{0.015}=0.563M$