# I need help with this question it relates to blood volume in a medical setting?

## You inject a 160 Ib male patient with 1.0 mL of a saline solution containing a radioactive form of sodium. It has an activity of 5.0 x 10^4 dpm. After allowing sufficient time for the solution to mix, you remove a 1.0 ml sample of blood and measure its radioactivity. You discover it to have an activity of 11 dpm. What is the patient's blood volume

May 24, 2016

$\approx 4.544 \text{ L}$

#### Explanation:

Let the blood volume of patient be $= n \text{ mL}$
After injecting saline solution new volume of blood $= n + 1 \text{ mL}$
DPM of $1 \text{ mL}$ injected sample$= 5.0 \times {10}^{4}$

Assuming there is no change in activity of sample during the time when two mix together.
DPM of $1 \text{ mL}$ extracted sample$= \frac{5.0 \times {10}^{4}}{n + 1}$ .....(1)
Measured DPM of $1 \text{ mL}$ extracted sample$= 11$ ......(2)
Equating (1) and (2)

$\frac{5.0 \times {10}^{4}}{n + 1} = 11$
$\left(n + 1\right) \times 11 = 5.0 \times {10}^{4}$, solving for $n$
$11 n = 5.0 \times {10}^{4} - 11$
$n = 4544. \overline{45} \text{ mL}$
$n \approx 4544 \text{ mL}$