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3 years ago

A sample of pure oxalic acid (H2C2O4.2H2O) weighs 0.2000 g and requires 30.12 ml of KOH solution for

Chemistry

1 answer:

ValentinkaMS [17]3 years ago

5 0

The molarity of KOH is 0.1055 M

calculation

Step 1: write the equation for reaction between H₂C₂O₄.2H₂O and KOH

H₂C₂O₄.2H₂O + 2 KOH → K₂C₂O₄ +4 H₂O

step 2: find the moles of H₂C₂O₄.2H₂O

moles = mass÷ molar mass

from periodic table the molar mass H₂C₂O₄.2H₂O= (1 x2) +(12 x2) +(16 x4) + 2(18)=126 g/mol

= 0.2000 g ÷ 126 g/mol =0.00159 moles

step 3: use the mole ratio to calculate the moles of KOH

H₂C₂O₄.2H₂O : KOH is 1:2

therefore the moles of KOH =0.00159 x 2 = 0.00318 moles

step 4: find molarity of KOH

molarity = moles/volume in liters

volume in liters = 30.12/1000=0.03012 L

molarity is therefore = 0.00318/0.03012 =0.1055 M

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