1.

A sample contains a mixture of NaHCO_(3) and Na_(2)CO_(3).HCl is added to 15,0 g. of the sample, yielding 11.0 g of NaCl. What percent of the sample is Na_(2)CO_(3) [("Reaction are :"),(Na_(2)CO_(3)+2HCl rarr 2 NaCl+CO_(2)+H_(2)O),(NaHCO_(3)+HCl rarr NaCl+CO_(2)+H_(2)O)] MW of NaCl=58.5, MW of NaHCO_(3)=84, MW of Na_(2)CO_(3)=106 g mol^(-1)

Answer»


Solution :Use `M=("% by WEIGHT "xx10xxd)/(Mw_(2))`
MOL. WT. of `C_(2)H_(5)OH=46 GM`
`M_(1)V_(1)=M_(2)V_(2)`
`(90xx10xx0.8)/46 xxV=(10xx10xx0.9)/46 xx40`
`V=(10xx0.9xx40)/(90xx0.8)=5 mL`.


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