A water sample is found to contain 96 ppm of $S O_{4}^{2 -}$ ion and 122 ppm of $H C O_{3}^{-}$ ion with $C a^{+ 2}$ as the only cation (Assume no hydrolysis of any ion).
I) The molarity of $S O_{4}^{2 -}$ ion in the sample is $10^{- P} M$ .
II) This water contains Q ppm $C a^{+ 2}$ .
III) 1 mol. of CaO is consumed in removing $H C O_{3}^{-}$ ion from $10^{R} g m$ of water described above. Determine the value of $\frac{P \times Q}{10 \times R}$ .

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answer is 4.

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Detailed Solution

P = 3, Q = 80, R = 6
I) $M_{S O_{4}^{2 -}} = \frac{1}{10^{6}} \times 1000 = 10^{- 3} M$
II) $n_{S O_{4}^{2 -}} = 1_{m o l}; n_{H C Q_{3}} = 2_{m o l}$
So, $n_{C a^{2 +}} = 1 + 1 = 2_{m o l}$ or $C a^{2 +} = 80 p p m$
III) $C a O + \underset{1 m o l}{C a {(H C O_{3})}_{2}} \to 2 C a C O_{3} + H_{2} O$
So, $n_{C a O} = 1 m o l$ $\frac{P \times Q}{10 R} = \frac{80 \times 3}{10 \times 6} = 4$