A sample of water has been analyzed for common ions and results are presented in the form of a bar diagram as shown.

The non-carbonate hardness (expressed in mg/L as $C a C O_{3}$ ) of the sample is

40

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205

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165

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195

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Solution

The correct option is A $40$
Total hardness
$= m e q / L o f C a^{2 +} a n d M g^{2 +}$

$= 2.65 + (4.1 - 2.65)$

$= 4.1 m e q / L$

$= 4.1 \times 50 = 205 m g / L a s C a C O_{3}$

Alkalinity
$= 3.3 m e q / L o f H C O_{3}^{-}$
$= 3.3 \times 50 = 165 m g / L o f C a C O_{3}$

Non-carbonate hardness
$= 205 - 165$
$= 40 m g / L o f C a C O_{3}$

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\begin{matrix} Ion & Concentration m g / l & Atomic Weight C a^{2 +} & 60 & 40 M g^{2 +} & 30 & 24.31 H C O_{3}^{-} & 400 & 61 \end{matrix}

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