InterviewSolution
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InterviewSolution
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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?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 gC. 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. |
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Answer» Correct Answer - A::D 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)` |
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