Which among the following has the highest normality? Options: A . &nbsp8 g of KOH per 100 ml of solution B . &nbsp0.5 M sulphuric acid C . &nbsp1 N phosphoric acid D . &nbsp2 moles of NaOH per 1000 ml of solution

Answer: Option D : D N = no.of equivalents Volume of solution(L) N of KOH = [ 8 56 ] 0.1 = 1.43 N N of H 2 S O 4 = M × n factor = 0.5 × 2 = 1 N equivalents of NaOH = moles × n factor = 2 × 1 = 2 eq N of NaOH = 2 1 = 2 N

Which among the following has the highest normality?

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 = \frac{no.of equivalents}{Volume of solution(L)}

N of KOH =

\frac{[\frac{8}{56}]}{0.1} = 1.43 N

N of

H_{2} S O_{4}

= M

\times

n factor =

0.5 \times 2 = 1 N

equivalents of NaOH = moles

\times

n factor
=

2 \times 1 = 2

eq
N of NaOH =

\frac{2}{1} = 2

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