Question

11 gm of CaCl2 is present in one litre of a hard water. Determine its hardness in degree clark.​

Answer 1

Answer:

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Explanation:

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Answer 2

Explanation:

For the given 11gm of cacl2 in one liter is calculated as follows:

the equation for calculating the hardness is given as

mass of hardness of producing substance × chemical equivalent of CaCo3 ÷ chemical equivalence hardness of producing substance

i.e.

mass of hardness of producing substance is 11 gm of CaCl2.

chemical equivalent of CaCo3 is 50.

chemical equivalence hardness of producing substance is 55.5

11 × 50 ÷ 55.5

=9.90 ppm or gm/L

for Degree Clark conversion

one ppm is equal to 0.07 degree Clark

therefore  for 9.90 ppm

9.09 × 0.07

=0.6363 degree Clark.

hence the hardness of water for 11gm of CaCl2 is 0.6363 degree Clark.