Question

How much volume of NaoH solution N/5 is required to neutralise 40 gm of oxalic acid

Answer 1

Answer:

Mass of NaOH required = 50.6 g

Explanation:

GIVEN :

 Normality of NaOH = N/5

Mass of oxalic acid = 40 gm

TO FIND : volume of NaOH required to neutralize oxalic acid

Step by Step Explanation :

reaction involved = 2NaOH + C₂H₂O₄  →    C₂O₄Na₂  + 2H₂O

Here , 1 mole of C₂H₂O₄ reacts with 2 moles of NaOH

mass of oxalic acid given = 40 g

so, no. of moles of oxalic acid = 40/63  where, 63 = molar mass of oxalic acid  ,

                  n (oxalic acid ) = 0.63

∴ no. of moles of NaOH = 2 x 0.63

                                       = 1.26

so, mass of NaOH required to neutralise oxalic acid = n x molar mass of NaOH

   =  1.26 x 40

mass of NaOH required  =  50.4 g

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