1. Given data

• Initial injected activity: $A_0 = 2.4\times 10^5\ \text{Bq}$
• Half-life of ${}^{131}\mathrm{I}$: $T_{1/2}=8\ \text{days}$
• Time after injection when blood is sampled: $t=11.5\ \text{hours}$
• Sample volume: $V_s=2.5\ \text{mL}$
• Activity of this sample: $A_s=115\ \text{Bq}$

Since the serum gets uniformly distributed in less than half an hour, we can treat the radioactive iodine as uniformly mixed in the blood when the sample is taken.

2. Find decay constant

Using $\lambda = \frac{\ln 2}{T_{1/2}}$

Convert half-life into hours: $T_{1/2}=8\times 24=192\ \text{hours}$

So, $\lambda = \frac{0.7}{192}\ \text{hour}^{-1}$

3. Activity remaining after 11.5 hours

Radioactive decay law: $A(t)=A_0 e^{-\lambda t}$

Thus, $A(t)=2.4\times 10^5\, e^{-\lambda t}$

Now, $\lambda t = \frac{0.7}{192}\times 11.5$

$\lambda t \approx 0.042$

Since this is small, use $e^{-x}\approx 1-x$

Hence, $e^{-0.042}\approx 1-0.042=0.958$

Therefore, $A(t)\approx 2.4\times 10^5\times 0.958$ $A(t)\approx 2.30\times 10^5\ \text{Bq}$

This is the total activity present in the whole blood volume at the time of sampling.

4. Use uniform distribution to find total blood volume

If total blood volume is $V$ mL, then activity per mL is $\frac{A(t)}{V}$

So activity in $2.5$ mL is $A_s = \frac{A(t)}{V}\times 2.5$

Given $A_s=115$ Bq: $115 = \frac{2.30\times 10^5}{V}\times 2.5$

Thus, $V=\frac{2.30\times 10^5\times 2.5}{115}$

$V=2.30\times 10^5\times \frac{1}{46}$

$V\approx 5000\ \text{mL}$

Therefore, $V \approx 5\ \text{L}$

5. Final answer

The total volume of blood in the person's body is approximately $\boxed{5\ \text{L}}$

6. Comparison with stored correct answer

Stored correct answer = $5$

Our derived answer = $5$

So they agree.