A water sample contains 81 ppm of $C a (H C O_{3})_{2}$ , 73 ppm of $M g (H C O_{3})_{2}$ , 68 ppm of $C a S O_{4}$ and 60 ppm of $M g S O_{4}$ . What is the hardness of the Water sample? (Atomic weights of calcium, magnesium and sulphur are respectively 40, 24 and 32).

50 ppm

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100 ppm

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150 ppm

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200 ppm

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Solution

The correct option is B 200 ppm
We can calculate the hardness of given water sample as:
(i) $C a (H C O_{3})_{2} \equiv C a C O_{3}$
$162 p p m \equiv 100 p p m$
$∴ 81 p p m \equiv 50 p p m$

(ii) $M g (H C O_{3})_{2} \equiv C a C O_{3}$
146 ppm $\equiv$ 100 ppm
$∴ 73 p p m \equiv 50 p p m$

(iii) $C a S O_{4} \equiv C a C O_{3}$
$136 p p m \equiv 100 p p m$
$∴ 68 p p m \equiv 50 p p m$

(iv) $M g S O_{4} \equiv C a C O_{3}$
$120 p p m \equiv 100 p p m$
$∴ 60 p p m \equiv 50 p p m$

The hardness is most commonly expressed as milligrams of calcium carbonate equivalent per litre. Calcium carbonate is the compound in hard water that causes limescale build-up. Soft water generally has less than 50 ppm of calcium carbonate, and hard water has over 200 ppm.

Total degree of hardness of water $= 50 \times 4 = 200 p p m$

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