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`        WHEN 100 ml OF M/10 H2SO4 IS MIXED WITH 500 ml OF M/10 NAOH THEN NATURE OF RESULTING SOLUTIONAND NORMALITY OF EXCESS OF REACTANT LEFT IS `
one year ago

Arun
24740 Points
```							This mixture would lead to a neutralization reaction as follows:H2SO4 + 2 NaOH → Na2SO4 + 2 H2OThus for every mole of H2SO4 two moles of NaOH are used.If u see, H2SO4 is limiting reagent.Thus, all H2SO4 will be used up while 0.03 (0.05–0.02) moles of NaOH are left.Thus solution would be basic.Normality of NaOH is=Equivalents of NaOH left(=moles for NaOH)/ Volume(L)=0.03/0.6 (Volume = 500ml + 100ml = 600ml = 0.6L)Normality = 1/20 N = 0.05 N
```
one year ago
Apurva Sharma
196 Points
```							n equivalent (n eq) = molarity x volumen eq of H2SO4 = 0.1 x 100 =10n eq of NaOH = 0.1 x 500 =50 now n eq of NaOH > H2SO4 i.e 50>40thus the resultant solution will be BASIC now Normality: NR= NbVb -NaVa /Vb+Va                               = 0.1x500-0.1x100/ 500+100                             =50-10/600                             =40/600 = 0.667N  hope this helps, give a thumbsup!
```
one year ago
Rishi Sharma
505 Points
```							Dear Student,Please find below the solution to your problem.Gram equivalent of h2so4=normality *volume(in litre)=1/5×100/1000=0.02Gram eq. Of Naoh=1/10×500/1000=0.05So 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=0.05NThanks and Regards
```
16 days ago
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