$10^{5}$ g of a sample of water contained 16.2 g of $C a (H C O_{3})_{2}$ . It's hardness is

200 ppm

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300 ppm

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60 ppm

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100 ppm

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Solution

The correct option is D 100 ppm
$10^{5} g H_{2} O - 16.2 g C a (H C O_{3})_{2}$
$10^{6} g H_{2} O - ?$
$\frac{10^{6} \times 16.2}{10^{5}} = 162 p p m o f C a (H C O_{3})_{2}$
$162 p p m C a (H C O_{3})_{2} = 100 p p m C a C O_{3}$

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What will be the hardness of $10^{5}$ grams water sample containing 16.2 grams of Ca( ${H C O}_{3}$ ) $_{2}$ ?

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10

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