A 5.00 mL sample was diluted to 1.000 L in a volumetric flask. The ethanol in a

Question

A 5.00 mL sample was diluted to 1.000 L in a volumetric flask. The ethanol in a 25.00 mL aliquot of the diluted solution was distilled into 50.00 mL of 0.02000 M K_2Cr_2O_7 and oxidized to acetic acid with heating as follows: 3C_2H_5OH + 2Cr_2O_7^2- + 16H^+ rightarrow 4Cr^3+ + 3CH_3COOH + 11H_2O After cooling, 20.00 mL of 0.1253 M Fe^2+ was pipetted into the flask for the following reaction: The excess Fe^2+ was then titrated with 7.46 mL of the standard K_2Cr_2O_7 to an indicator-based end point. Calculate the percent (w/v) ethanol (46.07 g/mol) in the sample.

Explanation / Answer

Solution :-

20.00 ml 0.1253 M Fe^2+

50.00 ml 0.0200 M K2Cr2O7

7.46 ml K2Cr2O7

Lets first calculate the moles of the K2Cr2O7

Moles = molarity * volume in liter

           = 0.0200 mol per L * 0.00746 L

           = 0.000149 mol

Now lets calculate the moles of the

Moles of the K2Cr2O7 in 50 ml solution = 0.0200 mol per L * 0.050 L = 0.001 mol

So the moles of the K2Cr2O7 reacted = 0.001 mol – 0.000149 mol = 0.000851 mol

Now lets calculate the moles of the ethanol reacted

0.000851 mol K2Cr2O7 * 3mol ethanol / 2 mol K2Cr2O7 = 0.001276 mol ethanol

Now lets calculate the mass of the ethanol

Mass = moles * molar mass

       = 0.001276 mol * 46.07 g per mol

         = 0.0588 g ethanol

So the w/v of ethanol = (0.0588 g / 5.00 ml )*100% = 1.18 %

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