Asample of pure oxalic acid (h2c2o4.2h2o) weighs 0.2000 g and requires 30.12 ml of koh solution for complete neutralization. calculate the molarity of the koh solution. (0.105 m)

 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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i do not get it   at all