Question

10 l of hard water required 0.56 g of lime (cao) for removing hardness. Hence, temporary hardness in ppm (part per million, ) of is:

Answer 1

hence there are 100 grams of Caco3 per 100 liters of water

Answer 2

Answer:

100

Explanation:

mass of Ca0= Ca + O

                    = 20+16 = 56g

given 0.56 Grams in Question

Thus, No. of mole = 0.56/56 = 0.01

Mass of Caco3= mole * eq wt

                        = 0.01 * 100 (mass of caco3)

                        = 1g

volume = 10*1000 =10000ml

hence ppm of 1g = 1g/10*0000 * 10^6 (ppm=10^6)

this comes out to be 100 on solving

Thank you :)