Answer:
Normality of KOH = 0.1062 N
<em>Note: The given question is not clearly stated. A related question below is answered and explained;</em>
<em> What is the NORMALITY of a solution of KOH if 45.18 mL are required to neutralize 0.3g of pure oxalic acid H2C2O4 . 2H2O?</em>
Explanation:
Equation of the reaction: 2KOH + H₂C₂O₄.2H₂O --> K₂C₂O₄ + 4H₂O
molar mass 0f KOH = 56 g/mol; molar mass of H₂C₂O₄.2H₂O = 126 g/mol
Number of moles of in 0.3 g of H₂C₂O₄.2H₂O = mass/molar mass
number of moles = 0.3 g / 126 g/mol = 0.0024 moles
From the equation of reaction, 2 moles of KOH are required to neutralize 1 mole of H₂C₂O₄.2H₂O
Number of moles of KOH required to neutralize 0.0024 moles of H₂C₂O₄.2H₂O = 2 * 0.0024 = 0.0048 moles
Molar concentration of 45.18 mL solution of KOH containing 0.0048 moles of KOH = number of moles/volume (L)
molarity of KOH solution = (0.0048/45.18 mL) * (1000 mL/1 L) = 0.1062 mol/L
Normality = n * molarity
where is the number of moles of OH⁻ ions produces from mole of KOH = 1
Normality of KOH = 1 * 0.1062
Normality of KOH = 0.1062 N
