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A `5.00 g` sample of vinegar is titrated with `0.108 M NaOH`. If the vinger requires `39.1 mL` of the `NaOH` for complete reaction, the mass percentage of acetic acid `(CH_(3)CO_(2)H)` in the vinegar isA. 0.0605B. 0.0407C. 0.0786D. 0.0506 |
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Answer» Correct Answer - 4 The reaction is `CH_(3)CO_(2)H(aq.)+NaOH(aq.) rarr CH_(3)CObar(O)Na^(+)(aq.)+H_(2)O(l)` Convert the volume of NaOH to moles of NaOH using the molarity formula. Then we convert moles NaOH to moles `CH_(3)CO_(2)H` using the balanced chemical equation. Finally we covert moles `CH_(3)CO_(2)H` to grams `CH_(3)CO_(2)H`. Mass percentage of acetic acid in the vinegar `=("Mass of acetic acid")/("Mass of vinegar")xx100` `n_(NaOH)=V_(NaOH "soln")xxM_(NaOH "soln")` `=(39.1xx10^(-3)L)(0.108 mol L^(-1))` `=4.22xx10^(-3) mol` According to equation, moles of `CH_(3)CO_(2)H` and `NaOH` reacting are equal. Thus `n_(CH_(3)COOH)=n_(NaOH)=4.22xx10^(-3) mol` `mass_(CH_(3)CO_(2)H)=n_(CH_(3)CO_(2)H)xx "molar mass"_(CH(3)CO_(2)H)` `=(4.22xx10^(-3) mol)(60 g mol^(-1))` `=0.253 g` Thus mass% `=(0.253 g)/(5.00 g)xx100%=5.06%` |
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