Question

The alcohol content in a $10.0$ g sample of blood from a driver required $4.23$ mL of $0.07654$ M $K_{2}C{r}_2{O}_7$ for titration. The mass $\%$ of alcohol is $x\times {10}^{-3}$. Here, $x$ is :
$3C{H}_{3}C{H}_{2}OH+2{C{r}_2{O}_7}^{2-}+16H^{+}\to 3C{H}_{3}COOH+4C{r}^{3+}+11H_{2}O$

Answer 1

$3$ moles ethanol $\equiv 2$ moles dichromate ion.
The number of moles of dichromate ion $=\frac{4.23}{1000}\times 0.07654M=3.23\times {10}^{-4}moles$.
The number of moles of ethanol $=\frac{3}{2}3.23\times {10}^{-4}moles=4.9\times {10}^{-4}moles$.
Mass of alcohol $=46\times \mathrm{4..9}\times {10}^{-4}=0.02234$ g.
Mass percentage $=\frac{0.0223g}{10.0g}\times 100=0.223=223\times {10}^{-3}\%$.
It is above $0.1\%$ prescribed limit.