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 ppm `CaCO_(3)` the amount of `Na_(2)CO_(3)` requried to soften 10 L of hard water is 10.6 g.
B. If hardness is 100 ppm `CaCO_(3)`, the amount of `Na_(2)CO_(3)` required to soften 10 L of hard is 10.6 g
C. If hardness is 420 ppm `MgCO_(3)`, the amount of `Na_(2)CO_(3)` required to soften 10 L of hard water is 53.0 g.
D. If hardness is 420 ppm `MgCO_(3)` 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=100gmol^(-1)`
Mw of `Na_(2)CO_(3)=46+60=106 g mol^(-1)`
(i). `underset(x mol)(CaSO_(4))+underset(x mol)(Na_(2)CO_(3))toCaCO_(3)+Na_(2)SO_(4)` ..(i)
(ii). `underset(y mol)(CaCl_(2))+underset(y mol)(Na_(2)CO_(3))toCaCO_(3)+2NaCl` .(ii)
(iii). 100 ppm `CaCO_(3)-=100g CaCO_(3)` in `10^(6)mL`
`=(100)/(100)mol` in `10^(6)mL`
Moles of `Na_(2)CO_(3)` required`=" mol of "CaCO_(3)`
`-=(100)/(100)mol` in `10^(6)mL`
`-=1 m" mol of "10^(6)mL`
`-=(1)/(10^(6))xx10xx10^(3)mL(10L)`
`=1xx10^(-2) mol` in `10L`
Therefore, moles of `Na_(2)CO_(3)` required
`=1xx10^(-2) mol` in `10L`
`=1xx10^(-2)xx(106g)/(10L)`
`=1.06g`
Similarly for 100 ppm `MgCO_(3):(Mw of MgCO_(3)=24+60=84gmol^(-1))`
(iv). `underset(xmol)(MgSO_(4))+underset(xmol)(Na_(2)CO_(3))toMgCO_(3)+Na_(2)SO_(4)`. (i)
(v). `underset(y mol)(MgCl_(2))+underset(y mol)(Na_(2)CO_(3))toMgCO_(3)+2NaCl` ..(ii)
(vi). `420 ppm MgCO_(3)-=420g MgCO_(3)` in `10^(6)` mL
`-=(420)/(84)=5 mol` in `10^(6)mL`
Moles of `Na_(2)CO_(3)` required
`-=" mol of "MgCO_(3)`
`-=5 mol` in `10^(6)mL`
`-=(5)/(10^(6))xx10xx10^(3)mL(10L)`
`-=5xx10^(-2) mol in 10L`
Moles of `Na_(2)CO_(3)` required `-=5xx10^(-2)mol(10L)^(-1)`
`-=5xx10^(-2)xx106g(10L)^(-1)`
`=(5.3g)/(10L)`