The hardness of 10,000 Litres of hard water sample was completely removed by passing it through a zeolite softener. The zeolite softener required 5000litres of NaCl solution containing 1170 mg of NaC

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written 3.1 years ago by

aganaka834 • 20

modified 2.9 years ago by

Krishna_Agrawal • 2.7k

engineering chemistry

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written 2.9 years ago by

Krishna_Agrawal • 2.7k

• modified 2.9 years ago

Given,

$V_1 = 10,000\ litres$

$V_2 = 5000\ litres$

$m = 1170\ mg = 1.17\ gm$

$We\ have\ to\ calculate\ the\ required\ hardness.$

$\implies Hardness (H) = \frac{50 \times m \times V_2 \times 10^3}{58.5 \times V_1}$

$\implies Hardness (H) = \frac{50 \times 1.17 \times 5000 \times 10^3}{58.5 \times 10,000} = 500\ ppm$

$\implies \therefore\ Hardness (H) = 500\ ppm\ Ans.$

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