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Concentration Terms - An Introduction (w/w etc)

One litre sam...

Question

$O n e l i t r e$ sample of water contains $0.9 m g$ of $N a C l_{2}$ and $0.9 m g$ of $M g C l_{2}$ . Find the total hardness in terms of parts permillion of $C a C O_{3}$ .

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Solution

$111 g C a C l_{2} \equiv 100 g C a C O_{3}$
$∴ 0.9 m g C a C l_{2} \equiv \frac{100}{111} \times 0.9 m g C a C O_{3}$
$\equiv 0.81 m g$ of $C a C O_{3}$
$95 g M g C l_{2} \equiv 100 g C a C O_{3}$
$∴ 0.9 m g M g C l_{2} \equiv \frac{100}{95} \times 0.9 m g C a C O_{3}$
$\equiv 0.94 m g$ of $C a C O_{3}$
Thus, $1$ litre waterr of $10^{6}$ mg water contains $(0, 81 + 0.94 = 1.75 m g)$ i.e., hardness of water is $1.75 p p m$ .

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Concentration Terms - An Introduction (w/w etc)

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