Question

The alcohol content in a $10.0$ g sample of blood from a driver required $4.23$ mL of $0.07654$ $M$ $K_2{Cr}_2{O}_7$ for titration. Should the police prosecute the individual for drunken driving?

• A. Yes
• B. No
• C. Nothing can be predicted
• D. None of the above

Answer 1

Answer: (A)

Yes

${3CH}_3{CH}_{2}OH+2K_2{Cr}_2{O}_7+8H_{2}S{O}_4⟶3{CH}_{3}COOH+2{Cr}_2{\left(S{O}_4\right)}_3+{2K}_{2}S{O}_4+11H_{2}O$

According to given data,

$4.23\times 0.07654$ millimoles of $K_2{Cr}_2{O}_7$$\equiv 4.23\times 0.07654\times 6$ milliequivalents of $K_2{Cr}_2{O}_7$

At equivalent point, alcohol will have same number of equivalents $\equiv 4.23\times 0.07654\times 6$ milliequivalent of ${CH}_3{CH}_{2}OH$

$\equiv \frac{4.23\times 0.07654\times 6\times 11.5}{1000}$ g ${CH}_3{CH}_{2}OH$ (as equivalent weight of alcohol is $\frac{46}{4}=11.5$)

$\equiv 0.022$ g in $10$ g sample

Thus, % of alcohol content in sample $=\frac{0.022}{10}\times 100$ $=0.22\ge 0.10$ (as current legal limit of blood alcohol content is 0.1 per cent by mass).

Police should prosecute the individual for drunken driving.

Hence, the correct option is $\text{A}$