Question

The following redox reaction occurs betwen the dichromate ion and ethanol:

3CH3CH2OH + 2CrO7^2- + 16H+ => 3CH3COOH + 4Cr^3+ + 11H2O
A chemist uses 26.25 mL of 0.500 M CrO7^2- to titrate a 10.0 mL sample of wine to the equivalence point.
a) What is the [CH3CH2OH] in the wine?
b) The concentration of ethanol in alcoholic beverages is expressed as percent by volume. If a wine is 10% alcohol, it means that there are 10 mL of ethanol for every 100mL of the beverage. The density of ethanol is 0.789 g/mL. Convert your answer in part (a) into percent by volume

Answer 1

a) 1,69 M

b)11.55%

Step-by-step explanation:

a) To determine the [CH3CH2OH] in the wine, we can use the stoichiometry of the balanced equation provided. Since the ratio of CrO7^2- to CH3CH2OH is 2:3, for every 2 moles of CrO7^2- that are consumed, 3 moles of CH3CH2OH are produced. Therefore, we can use the volume and molarity of the CrO7^2- solution to determine the number of moles of CH3CH2OH present in the wine sample.

The number of moles of CrO7^2- used in the titration is (26.25 mL)(0.500 M) = 13.125 moles. Therefore, the number of moles of CH3CH2OH present in the wine sample is (3/2)(13.125 moles) = 19.69 moles

Since the original volume of wine sample was 10.0 mL, the [CH3CH2OH] in the wine is (19.69 moles) / (0.01 L) = 1.969 M

b) To convert the answer in part (a) into percent by volume, we can use the density of ethanol and the conversion factor of 100mL/1L.

The density of ethanol is 0.789 g/mL, so the mass of ethanol in the sample is (19.69 moles) x (46.07 g/mol) = 910.1 g

The mass of ethanol in 100ml is (910.1g) x (100mL/1L) = 91.01 g/100mL

The volume of ethanol in the sample is (91.01 g/100mL) / (0.789 g/mL) =115.5 mL/L or 11.55% by volume.