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Concentration Terms - An Introduction (w/w etc)

Solution of ...

Question

Solution of $100 m l$ water contains $0.73 g$ of $M g (H C O_{3})_{2}$ and $0.81 g$ of $C a (H C O_{3})_{2}$ . Calculate the hardness in terms of ppm of $C a C O_{3}$ .

10^{2} p p m

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10^{4} p p m

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5 \times 10^{3} p p m

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10^{3} p p m

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Solution

The correct option is D $10^{4} p p m$
$p p m$ of $C a C O_{3} = \frac{(\frac{0.73}{146} + \frac{0.81}{162}) \times 100}{100} \times 10^{6} = 10^{4} p p m$

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