Presence of which makes water unsuitable for drinking.

Fe = 0.2 ppm

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Cu = 2 ppm

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Mn = 0.5 ppm

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Zn = 0.05 ppm

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Solution

The correct option is C
Mn = 0.5 ppm

Explanation of correct answer:-
Option (C):
Presence of Manganese ions more than 0.5 ppm causes discoloration
It makes unpleasant taste in the water which makes it unsuitable for drinking purpose.
Plenty of Manganese causes toxicity in the body.
It reduces memory, motor skills causing nausea, constipation, etc.
Therefore, option (C) is the correct answer.

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Similar questions

F^{-} = 10, S O_{4}^{2 -} = 100, N O_{3}^{-} = 50

F^{-} = 10; S O_{3}^{2 -} = 100; N O_{3}^{-} = 50

F e = 0.2; M n = 5.0; C u = 3.0; Z n = 5.0

A water sample has $ppm$ level concentration of following anions

$F^{-} = 10; {SO}_{4}^{2 -} = 100; {NO}_{3}^{-} = 50$

The anion/ anions that makes the water sample unsuitable for drinking is/are:

p p m

F^{-} = 10, S O_{4}^{2 -} = 100, N O_{3}^{-} = 50

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