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ID: 712452 • Letter: M
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Mailings Review View Help FORT POF Tell me what erences 1 No Spac... Heading 7 Subtle Em... Erm Styles Paragraph 15. The alcohol content in a 10.0 g sample of blood from a driver required 4.23 mL of 0.07654 M K Cr:0, for titration. If the current legal limit of blood alcohol content is 0.08 percent by mass, should the driver be prosecuted for drunken driving? What is the percent alchohol in the driver's blood? (7 points) 3 CllsOH (ethanol) + 2 Kort): + 8H2SOa 3 HC1102 + 2 Cr:(SO4)s +2 K:S04 + 11 H20 (Reference: Chang 4.154) 16. Write balanced chemical equations for the following: (8 points) a. A solution of sulfuric acid (battery acid) is neutralized with a solution of sodium hydroxide. b. The combustion of iron metal produces iron(III) oxide (rust). c. When sodium metal is dropped into water, it fizzes. d. The science fair "volcano reaction" is produced by mixing sodium bicarbonate (baking soda) with vinegar (a solution of acetic acid HC2HsO2). Copyright © 2017 by Thomas Edison State University. All rights reserved o eExplanation / Answer
15.
no. of mole = molarity X volume of solution in liter
4.23 ml = 0.00423 liter
no. of mole K2Cr2O7 used for titration = 0.07654 X 0.00423 = 0.0003237642 mole
According to reaction 2 mole of K2Cr2O7 react with 3 mole of C2H5OH therefore to react with 0.0003237642 mole of K2Cr2O7 required C2H5OH = C2H5OH X 3 / 2 = 0.0004856463 mole
molar mass of C2H5OH = 46.07 g/mol then 0.0004856463 mole of C2H5OH = 0.0004856463 X 46.07 = 0.022373725 gm
C2H5OH present in blood sample = 0.022373725 gm
10 gm blood sample = 100 % then 0.022373725 gm = 0.022373725 X 100 / 10 = 0.224 %
Blood sample contain 0.224 % alcohol legal limit is 0.08 % therefore driver is drunken.
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