The hardness of water sample (in terms of equivalents of $\ce{CaCO3}$) containing $10^{-3}$ M $\ce{CaSO4}$ is (molar mass of CaSO$_4$ = 136 g mol$^{-1}$)

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Hydrogen

The hardness of water sample (in terms of equivalents of

CaCO X_{3}

) containing

1 0^{- 3}

CaSO X_{4}

is
(molar mass of CaSO

_{4}

= 136 g mol

^{- 1}

)

Answer: (D)

Hardness is quoted as the mass of

CaCO X_{3}

that would give the same effect, per million parts of water. One mole of

CaSO X_{4}

corresponds to one mole of

CaCO X_{3}

, so a

1 0^{- 3} M

solution carries

1 0^{- 3}

mol of

CaCO X_{3}

-equivalent per litre. That is

1 0^{- 3} \times 100 = 0.1 g

CaCO X_{3}

in a litre of water weighing about

1000 g

. Expressed as parts per million,

\frac{0.1}{1000} \times 1 0^{6} = 100 ppm

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option