$3.95 g$ of $KMnO X_{4}$ (molar mass $158 g mo l^{- 1}$ ) is dissolved in water to make $250 mL$ of solution. Select the correct statement(s).

AIts normality as an oxidant in acidic medium is $0.50 N$
BIts normality as an oxidant in neutral medium ( $MnO X_{2}$ formed) is $0.30 N$
CIts normality in strongly alkaline medium ( $MnO X_{4} X^{2 -}$ formed) is $0.50 N$
DThe molarity of the solution is $0.10 M$

Answer: A, B, D

Molarity

= 3.95/158/0.250 = 0.10 M

, independent of medium.

N = M \times n

: acidic

n = 5

gives

0.50

, neutral

n = 3

gives

0.30

, strongly alkaline

n = 1

gives

0.10

(not

0.50

). So options 1, 2 and 4 are correct.