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Q. Which amongst the following has the highest normality ?

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Solution:

$8 \,g \,KOH / 100cc=\frac{8}{56}\times\frac{1000}{100}=\frac{10}{7}N$
$=1.4\,N$,
$0.5 \,M\, H_{2}SO_{4}=1\,N \, H_{2}SO_{4}$
$6\, g \,NaOH/ 100cc=\frac{6}{40}\times\frac{1000}{100}$
$=1.5\,N$