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The hardness of a water sample (in terms of equivalents of $\text{CaCO}_3$) containing $10^{-3}$ M $\text{CaSO}_4$ is: (molar mass of $\text{CaSO}_4$ = 136 g mol$^{-1}$)
(a) 10 ppm (b) 50 ppm (c) 90 ppm (d) 100 ppm
$10^{-3}$ M $\text{CaSO}_4 \equiv 10^{-3}$ M $\text{CaCO}_3$. $W = 10^{-3} \times 100 = 100$ mg per litre = 100 ppm.
BETA
