Answered: The alcohol content in a 10.0 g sample…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

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

Expert Solution

Step 1

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,

$M = \frac{n}{V}$

where

M: molarity of K₂Cr₂O₇ in M.

V: Volume of K₂Cr₂O₇in liters.

n: number of moles of K₂Cr₂O_7.

Substituting all the values in the above equation as-

$\begin{array}{rcl} 0.07654 m o l / L & = & \frac{n}{4.23 \times 10^{- 3} L} \\ n & = & 0.07654 m o l / L \times (4.23 \times 10^{- 3}) L \\ n & = & 0.0003237 m o l \end{array}$

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