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What volume of 90% alcohol by weight (d=0.8 g mL^(-1)) of ethanol must be used to prepare 40 mL of 10% alcohol by weight (d=0.9 g mL^(-1))

Answer»


SOLUTION :Let x of `Na_(2)CO_(3)` . Then, weight of `NaHCO_(3)=(15-x)g`
Moles of `NaCl` is produced `=(11.0 g)/(58.5 g)=0.188` MOL
The `NaCl` is produced by the reaction of `(x/106)` mol of `Na_(2)CO_(3)` and `((15-x))/84` mol of `NaHCO_(3)`. Each mol of `Na_(2)CO_(3)` produces 2 mol of `NaCl`
`:. (2X)/106+(15-x)/84=0.188`
Solve `x := 1.35 g Na_(2)CO_(3)`,
`NaHCO_(3)=(15-1.35)=13.6 g`
`% Na_(2)CO_(3)=13.5/15xx100=9.0% Na_(2)CO_(3)`.


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