The alcohol content in a 10.0 g sample of blood from a driver required 4.23 mL of 0.07654 M K2Cr2O7 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 alcohol in the driver’s blood? 3 C2H5OH (ethanol) + 2 K2Cr2O7 + 8 H2SO4 → 3 HC2H3O2 + 2 Cr2(SO4)3 + 2 K2SO4 + 11 H2O
The alcohol content in a 10.0 g sample of blood from a driver required 4.23 mL of 0.07654 M K2Cr2O7 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 alcohol in the driver’s blood?
3 C2H5OH (ethanol) + 2 K2Cr2O7 + 8 H2SO4 → 3 HC2H3O2 + 2 Cr2(SO4)3 + 2 K2SO4 + 11 H2O
The volume of K2Cr2O7 = 4.23 mL
= 4.23×10-3 L
Molarity of K2Cr2O7 = 0.07654 mol/L(M)
Therefore, molarity is defined as the number of moles of solute per unit volume of solution in litres.
Mathematically,
where
M: molarity of K2Cr2O7 in M.
V: Volume of K2Cr2O7 in liters.
n: number of moles of K2Cr2O7.
Substituting all the values in the above equation as-
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