plz give the proper correct answer_____
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kishor5919:
plz solve this... very needed
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acidic , N/5 .. i am not sure about this answer though
plz give the process of this prblm
with a pic
Gram equivalent of h2so4=normality *volume(in litre)=1/5×100/1000=0.02
Gram eq. Of Naoh=1/10×500/1000=0.05
So nature of solution will be basic as gram eq of Naoh are more..
So after use of 0.02 eq H2so4, eq left=0.05–0.02=0.03 (600 ml/0.6 lit)
So normality of excess reactant left is=gram Eq/vol(lit.)=0.03/0.6=0.05N
Gram eq. Of Naoh=1/10×500/1000=0.05
So nature of solution will be basic as gram eq of Naoh are more..
So after use of 0.02 eq H2so4, eq left=0.05–0.02=0.03 (600 ml/0.6 lit)
So normality of excess reactant left is=gram Eq/vol(lit.)=0.03/0.6=0.05N
i ve just solved this .. is this correct ??
well the 1st answer i wrote of was wrong .. i am not sure about this one too
crrct yr
can u solve this in a rough papr? unable to understand
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