1.

`RH_(2)` ( ion exchange resin) can replace `Ca^(2+)`d in hard water as. `RH_(2)+Ca^(2+)rarrRCa+2H^(+)` `1 "litre"` of hard water passing through `RH_(2)` has `pH2`. Hence hardness in `pp m "of" Ca^(2+)` is:A. `200`B. `100`C. `50`D. `125`

Answer» Correct Answer - A
`[H^(+)]=10^(-2)`
`:. [Ca^(2+)]=(10^(-2))/(2)=(40xx10^(-2))/(2)=0.2 g//"litre"`
`:. "Hardness"=(0.2xx10^(6))/(10^(3))=200pp m`


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