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Hardness of Water and Treatment

10L of water ...

Question

$10$ L of water requires $0.28$ g of lime (CaO) for removing hardness. Calculate the temporary hardness in ppm of $C a C O_{3}$ .

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Solution

Given concentration of lime= $0.28 g = 280 m g / 10 l i t = 28 m g / l i t$
$1 p p m = 1$ part of $C a C O_{3}$ in $10^{6}$ parts of $H_{2} O$ . $[m = \frac{100}{28}] \to equivalent of C a O$
Temporary hardness= $c \times m = 28 \times \frac{100}{28} = 100$
$∴$ Temporary hardness= $100 p p m$ .

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