Permanent hardness is due to `CI^(ɵ)` and `SO_4^(2-)` of `Mg^(2+)` and `Ca^(2+)` and is removed by adding `Na_2CO_3`.
`{:(CaSO_(4)+Na_(2)CO_(3)toCaCO_(3)+Na_(2)SO_(4)),(CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)+2NaCl):}` Which of the following statements is`//`are correct?

Options

A. If hardness is `100 p p m CaCO_(3)` the amount of `Na_(2)CO_(3)` required to soften `10 L` of hard water is `10.6 g`
B. If hardness is `100 p p m CaCO_(3)`, the amount of `Na_(2)CO_(3)` required to soften `10L` of hard water is `1.06 g`.
C. If hardness is `420 p p m MgCO_(3)`, the amount of `Na_(2)CO_(3)`required to soften `10 L` of hard water is `53 g`
D. If hardness is `420 p p m MgCO_(3)` is the amount of `Na_(2)CO_(3)` required to soften `10 L` of hard water is `5.3 g`

Answer

A, D

Explanation

`MW of CaCO_(3)=40+12+3xx16=100 gmol^(-1)`
`MW of Na_(2)CO_(3)=46+12+3xx16=106 gmol^(-1)`
`{:(i.,CaSO_(4)+,Na_(2)CO_(3)to,CaCO_(3),+Na_(2)SO_(4),....(i)),(,x mol,xmol,-,-,),(,-,-,x mol,-,),(ii.,CaCl_(2),+Na_(2)CO_(3)to,CaCO_(3),+2NaCl,....(ii)),(,y mol,y mol,-,-,),(,-,-,y mol,-,):}`
iii. `100 p p m CaCO_(3) -=100 g CaCO_(3)` in `10^(6) mL`
`=100/100 mol `in `10^(6) mL`
Moles of `Na_(2)CO_(3)` required =moles of `CaCO_(3)`
`=100/100 mol` in `10^(6) mL` ltbRgt `1/(10^(6))xx10xx10^(3)mL (10 L)`
`=1xx10^(-2) mol ` in `10 L`
Therefore, moles of `Na_(2)CO_(3)` required
`=1xx10^(-2) mol ` in `10 L`
`=1xx10^(-2)xx106 g//10 L`
`=1.06 g`
Similarly, for `100 p p m MgCO_(3)`,
`MW of MgCO_(3)=24+12+16xx3=84 g mol^(-1)`
`{:(iv.,MgSO_(4)+,Na_(2)CO_(3)to,MgCO_(3),+Na_(2)SO_(4),....(i)),(,x mol,xmol,-,-,),(,-,x mol,,,),(v.,MgCl_(2),+Na_(2)CO_(3)to,MgCO_(3),+2NaCl,....(ii)),(,y mol,y mol,-,-,),(,-,-,,y mol,):}`
vi. 420 p p m `MgCO_(3)-=420 g MgCO_(3) ` in `10^(6) mL`
-=420/84=5 mol in `10^(6) mL`
Moles of `Na_(2)CO_(3)` required =`5xx10^(-2) mol//10 L`
`=5.3 g //10 L`