Given,

$V_1 = 10,000\ litres$

$V_2 = 5000\ litres$

$m = 1170\ mg = 1.17\ gm$

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

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

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