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an aqueous solution of 6.3 gm oxalic acid dihydrate is made upto 250 ml. the volume of .1 N NaOH required to completely neutralise 10 ml of this solution is?? an aqueous solution of 6.3 gm oxalic acid dihydrate is made upto 250 ml. the volume of .1 N NaOH required to completely neutralise 10 ml of this solution is??
Molarity for oxalic acid is:M x 126 x 250 = 6.3 x 1000M = 0.2and normality would become hence, = 0.2 x 2 = 0.4.From N1 V1 = N2 V20.4 x 10 = 0.1 x VV = 40 ml is the req. answer.
oxalic acid given mass = 6.3gvolume of given solution= 250mlnormality of naoh=0.1N moles of C2H6O6(oxalic acid)= 6.3/126 = 0.05 (molarity*volume(in litre) = moles)so molarity*250=moles of oxalic acidM*250/1000 = 0.05 M=0.2 n factor of oxalic acid=2(normality=molarity*nfactor)so normality=0.4(normality*volume(in ml) = mili equivalents)so Meq=0.4*10=4(meq of acid = meq of base)so meq of naoh=4volume of naoh=40ml
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