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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??

Grade:11

2 Answers

Vikas TU
14149 Points
7 years ago
Molarity for oxalic acid is:
M x 126 x 250 = 6.3 x 1000
M = 0.2
and normality would become hence, = 0.2 x 2 = 0.4.
From N1 V1 = N2 V2
0.4 x 10 = 0.1 x V
V = 40 ml is the req. answer.
vignesh
32 Points
6 years ago
oxalic acid given mass = 6.3g
volume of given solution= 250ml
normality 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 acid
M*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=4
volume of naoh=40ml
 

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