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Find out the volume in mL of 0.1 N HCl solution required to react completely wth 1.0 g of a mixture of Na_(2)CO_(3) and NaHCO_(3)containing equimolar amounts of the two compounds . |
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Answer» Solution :Let the amount of `Na_(2)CO_(3)` in 1 G of mixture be X g and sincethe `Na_(2)CO_(3) and NaHCO_(3)`are in equimolar amounts , `x/106 = (1-x)/84 , " where " {{:("mol . WT . Of "Na_(2)CO_(3)=106),(" mol .wt of "NaHCO_(3)=84):}}` ` :. "" x = 0.558 g ` Thus wt. of `na_(2)CO_(3) = 0.558 g ` and wt. of `NaHCO_(3) = 1 - 0.558 = 0.442 g ` Now , m.e of HCL = m.e of `Na_(2)CO_(3) + " m.e of " NaHCO_(3)` m.e of HCl = eq. of `Na_(2)CO_(3) xx 1000 + " eq. of " NaHCO_(3) xx 1000 ""...(Eqn.3)` If v is the volume of HCl in mL then `0.1 xx v = (0.558)/53 xx 1000 + (0.442)/84 xx 1000 "" ...(Eqn. 4i)` ` :. "" v = 157 .9 mL ` |
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