Question -
Which among the following has the highest normality?
Options:
A .  8 g of KOH per 100 ml of solution
B .  0.5 M sulphuric acid
C .  1 N phosphoric acid
D .  2 moles of NaOH per 1000 ml of solution
Answer: Option D : D N=no.of equivalentsVolume of solution(L) N of KOH = [856]0.1=1.43N N of H2SO4 = M × n factor = 0.5×2=1N equivalents of NaOH = moles × n factor = 2×1=2eq N of NaOH = 21=2N
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