A) 10 ppm
B) 100 ppm
C) 50 ppm
D) 90 ppm
Correct Answer: B
Solution :
\[{{10}^{-3}}M\]means \[{{10}^{-3}}\]moles of\[CaS{{O}_{4}}\]present in 1 L of water. \[{{10}^{-3}}moles\,CaS{{O}_{4}}=\frac{Mass\,of\,CaS{{O}_{4}}}{Molar\,mass\,of\,CaS{{O}_{4}}}\] Mass of \[CaS{{O}_{4}}={{10}^{-3}}mol\times 136g\,mo{{l}^{-1}}\]or 136 mg i.e., \[\because \]136 mg of \[CaS{{O}_{4}}\]present in 1 kg of water \[\therefore \] \[{{10}^{6}}g\]of water will have \[136000mg\,CaS{{O}_{4}}\] \[136gCaS{{O}_{4}}\equiv 100g\,CaC{{O}_{3}}\] \[13600mg\,CaS{{O}_{4}}\equiv \frac{100}{136}\times \frac{136000}{1000}=100g\,CaC{{O}_{3}}\]Thus, hardness of water =100 ppm
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