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(D) Hardness is expressed in terms of $CaCO_3$ equivalents per $10^6$ parts of water $(ppm)$.Given concentration of $CaSO_4 = 10^{-3} \ M = 10^{-3} \

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$100$

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(D) Hardness is expressed in terms of $CaCO_3$ equivalents per $10^6$ parts of water $(ppm)$.Given concentration of $CaSO_4 = 10^{-3} \ M = 10^{-3} \ mol \ L^{-1}$.Since $1 \ mol$ of $CaSO_4$ provides $1 \ mol$ of $Ca^{2+}$ ions,the concentration of $Ca^{2+}$ is $10^{-3} \ mol \ L^{-1}$.Equivalents of $CaCO_3$ = Equivalents of $Ca^{2+}$.$n_{CaCO_3} 	imes v.f. = n_{Ca^{2+}} 	imes v.f.$$n_{CaCO_3} 	imes 2 = 10^{-3} 	imes 2 \implies n_{CaCO_3} = 10^{-3} \ mol \ L^{-1}$.Mass of $CaCO_3 = 10^{-3} \ mol 	imes 100 \ g \ mol^{-1} = 0.1 \ g \ L^{-1}$.Assuming the density of water is $1 \ g \ mL^{-1}$,$1 \ L$ of water weighs $1000 \ g$.Hardness in $ppm = \frac{	ext{mass of } CaCO_3 	ext{ in } g}{	ext{mass of water in } g} 	imes 10^6$.Hardness $= \frac{0.1 \ g}{1000 \ g} 	imes 10^6 = 100 \ ppm$.

The hardness of a water sample (in terms of equivalents of $CaCO_3$) containing $10^{-3} \ M \ CaSO_4$ is ................... $ppm$ (Molecular mass of $CaSO_4 = 136 \ g \ mol^{-1}$)

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