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Standard XII

Chemistry

Hardness of Water and Treatment

6 ml of a sta...

Question

$6$ ml of a standard soap solution $(1$ ml $= 0.001$ g $)$ of $C a C O_{3}$ were required in titrating $50$ ml of water to produce a good lather. Its degree of hardness is:

120

ppm

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100

ppm

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50

ppm

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60

ppm

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Solution

The correct option is A $120$ ppm
Degree of hardness $= \frac{6 \times 0.001}{0.05} = 120$ ppm.

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6 ml of a standard soap solution (1 ml =0.001 g) of CaCO3 were required in titrating 50 ml of water to produce a good lather. Its degree of hardness is:

The correct option is A 120 ppmDegree of hardness =6×0.0010.05=120 ppm.

$6$ ml of a standard soap solution $(1$ ml $= 0.001$ g $)$ of $C a C O_{3}$ were required in titrating $50$ ml of water to produce a good lather. Its degree of hardness is:

The correct option is A $120$ ppm
Degree of hardness $= \frac{6 \times 0.001}{0.05} = 120$ ppm.