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 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` |
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Answer» Correct Answer - A::D `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` |
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