NEET Chemistry Hydrogen Questions Solved

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50 lit. of hard water of 120 ppm temporary hardness is mixed with 1.62 kg of CaOH2, what is the hardness of the resulting water ? (The sp. gr. of hard water = 1 gm/ml)

(A) 8.088 × 105 ppm                               

(B) 8.088 × 104 ppm

(C) 80 ppm                                             

(D) 808 ppm

Concept Videos :-

#3 | Hard & Soft Water

Concept Questions :-

Hard & Soft Water

B.    The wt. of 50 lit. of hard water=50×103gm            wt. of CaCO3 present=120×50×103106gm            50 gm CaCO3=81 gm of CaHCO32            120×50×103106gm CaCO3            8150×120×50×103106gm CaHCO32            81×120×103 gm CaHCO32            9.72 gm CaHCO32            0.06 moles of CaHCO32             moles of CaOH2 added=1.62×10340=40.5             moles CaOH2 reacted =0.06                 moles CaOH2 remaining =40.5-0.06=40.44             moles Ca2+ in 50 lit.=40.44             moles CaCO3 in 50 lit.=40.44             amount of CaCO3 in 50 lit.=40.44×100=4044 gm             hardness or resulting water                =40445×104×106=8.088×104 ppm

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