1.42g of a mixture of caCO_(3) and MgCO_(3)was dissolved in 200 mL of0.2 N HCl solutionwhich was then diluted to 250 mL . 10 mL of this solution was neutralised by 12 mL of (N/30) Na_(2)CO_(3) .Find out the percentage of eachin the mixture. ( Ca = 40 , Mg = 24 , C = 12 , O = 16)

1.42g of a mixture of caCO_(3) and MgCO_(3)was dissolved in 200 mL of0

1.	1.42g of a mixture of caCO_(3) and MgCO_(3)was dissolved in 200 mL of0.2 N HCl solutionwhich was then diluted to 250 mL . 10 mL of this solution was neutralised by 12 mL of (N/30) Na_(2)CO_(3) .Find out the percentage of eachin the mixture. ( Ca = 40 , Mg = 24 , C = 12 , O = 16)
Answer» Solution :Let the wt. of `CaCO_(3)` be x G . ` :. ` wt . Of `MgCO_(3) = (1.42- x ) g ` ` :. ` eq of `CaCO_(3) = x/50` and eq. of `MgCO_(3) = (1.42 - x)/42 ""…(Eqn.4)` `( " eq. wt of " CaCO_(3) = 100/2 = 50 " eq. wt of " MgCO_(3) = 84/2 = 42)` Total m.e of `CaCO_(3) and MgCO_(3) = x/50 xx 1000 + (1.42 - x)/42 xx 1000 ...(Eqn.3)` m.e of HCl = `0.2 xx 200 = 40 ` From the givenquestion it is clearthat m.eof HCl is greater than those of `caCO_(3) and MgCO_(3)` `:. ` m.e of excess HCl = m.e of HCl - m.e of `CaCO_(3) and MgCO_(3)` ` = 40 - { 1000 (x/50 + (1.42 -x)/42 )}` ` :. ` the m.e of the resulting solution does not change on dilution . ` :. ` normality of excess HClin the diluted resulting solution ` = (m.e)/250 = (4 - {1000(x/50+(1.42-x)/42)})/250` ` :. ` m.e of 10 ML of the resulting solution ` (4 - {1000(x/50+(1.42-x)/42)})/250xx10` m.e of `na_(2)CO_(3)` solution ` = 1/30 xx12 ` ...(Eqn.1) ` :. (40 - {1000(x/50+(1.42-x)/42)})/250 xx 10 = 1/ 30 xx 12 "" ( Eqn. 2)` ` :. x = 1 ` ` :. % of CaCO_(3) = 1/(1.42) xx 100 = 70.4 % ` % of `MgCO_(3) = 100 - 70.4 = 29 .6 % `