91Ó°ÊÓ

As you know,

$n_{C{r}_2{O}_7^{2-}}=V_{C{r}_2{O}_7^{2-}}×M_{C{r}_2{O}_7^{2-}}$

Where,

$n_{C{r}_2{O}_7^{2-}}$= moles of Cr₂O₇^2-

$V_{C{r}_2{O}_7^{2-}}$= volume of Cr₂O₇^2-

$M_{C{r}_2{O}_7^{2-}}$= Molarity of Cr₂O₇^2-

According to the question;

$V_{C{r}_2{O}_7^{2-}}$ = 35.46mL = 0.03546L

$M_{C{r}_2{O}_7^{2-}}$ = 0.05961 mol/L

Now, you can write

$$\begin{gathered} n_{C{r}_2{O}_7^{2-}}=V_{C{r}_2{O}_7^{2-}}×M_{C{r}_2{O}_7^{2-}} \\ {n}_{C{r}_2{O}_7^{2-}}=0.03546×0.05961\left(mol\right) \\ {n}_{C{r}_2{O}_7^{2-}}=0.0021\left(mol\right) \end{gathered}$$

0.0021 mol of Cr₂O₇^2- were required to titrate 28g of plasma.

According to the balanced equation

$16H^{+}\left(aq\right)+2C{r}_2{O}_7^{2-}\left(aq\right)+C_2{H}_{5}OH\left(aq\right)→4C{r}^{3+}\left(aq\right)+2C{O}_2\left(g\right)+11H_{2}O\left(l\right)$

2 moles of Cr₂O₇^2- reacts with 1 moles of C₂H₅OH

Hence, moles of C₂H₅OH () can be calculated as:

$$\begin{gathered} n_{C_2{H}_{5}OH}=n_{C{r}_2{O}_7^{2-}}×\frac{1}{2} \\ {n}_{C_2{H}_{5}OH}=0.0021×\frac{1}{2}\left(mol\right) \\ {n}_{C_2{H}_{5}OH}=0.00105\left(mol\right) \end{gathered}$$

0.00105 moles of C₂H₅OH were present in the 28g sample of plasma.

As you know,

$m_{C_2{H}_{5}OH}=n_{C_2{H}_{5}OH}×M_{C_2{H}_{5}OH}$

Where,

$m_{C_2{H}_{5}OH}$= mass of C₂H₅OH

$n_{C_2{H}_{5}OH}$= moles of C₂H₅OH = 0.00105 mol

$M_{C_2{H}_{5}OH}$= molar mass of C₂H₅OH = 46.07 g/mol

Now you can write,

role="math" localid="1663305550168" $$\begin{gathered} m_{C_2{H}_{5}OH}=n_{C_2{H}_{5}OH}×M_{C_2{H}_{5}OH} \\ {m}_{C_2{H}_{5}OH}=0.00105×46.07\left(g\right) \\ {m}_{C_2{H}_{5}OH}=0.04837\left(g\right) \end{gathered}$$

0.04837 grams of C₂H₅OH were in the blood.

As you know,

Mass percent of C₂H₅OH

$=\frac{m_{C_2{H}_{5}OH}}{m_S}×100\%$

Where, m_S = mass of sample.

Now you can write

Mass percent of C₂H₅OH

$$\begin{gathered} =\frac{m_{C_2{H}_{5}OH}}{m_S}×100\% \\ =\frac{0.04837}{28}×100\% \\ =0.173\% \\ \backslash \mathrm{endgathered} \end{gathered}$$

The mass percentage of alcohol in the blood is 0.173%.