Grade 11Inorganic Chemistry

how to calculate degree of hardness of water?
please send as early as possible

chakri0712

9 Years agoGrade 11

5 Answers

Umakant biswal

9 Years ago

@ chakri
we cannot have any special formula for that , but we can find by the following as example
we need to calculate the hardness if 9.6 mg of Mgso4 present in 1 kg of water ,
so, here
given , 1000g of water contains Mgso4 = 9.6 mg
1000000g of water contain mgso4 = 9.6 * 1000 = 9600 mg = 9.6 g
u need to remember – 100000g = one million parts of water
1 mole of mgso4 = 1 mole of caco3
120 g of mgso4 = 100g of caco3
9.6 g of mgso4 = (100g * 9.6 g ) / 120 g of caco3 = 8g of caco3
so, here degree of hardness = 8 ppm
HOPE IT CLEARS

vishnu budarapu

9 Years ago

degree of hardness =(GMwt. of CaCO3 /GMwt. of given salt) X (wt of given salt / wt of taken water) X 1000000

saurabh singh

8 Years ago

degree of hardness=strength of hardness producing substance×multiplication factor so,multiplication factor=100/2×chemical equivalent of hardness producing substance.

Om agrawal

6 Years ago

We can find hardness of substance woth relative to CaCO3

Ex- hardness of mgso4 of 90 gram

120 gram MSO4=100 gram CaCO3

90 gram = 100*90/120

75 ppm [ ans]

Yash Chourasiya

6 Years ago

Dear Students

we cannot have any special formula for that , but we can find by the following as example
we need to calculate the hardness if 9.6 mg of Mgso4 present in 1 kg of water ,
so, here
given ,
1000g of water contains Mgso4 = 9.6 mg
1000000g of water contain mgso4 = 9.6 * 1000 = 9600 mg = 9.6 g
You need to remember : – 100000g = one million parts of water
1 mole of mgso4 = 1 mole of caco3
120 g of mgso4 = 100g of caco3
9.6 g of mgso4 = (100g * 9.6 g ) / 120 g of caco3 = 8g of caco3
so, here degree of hardness = 8 ppm

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya