Question

Creatinine, ${\mathbf{C}}_4{\mathbf{H}}_7{\mathbf{N}}_3\mathbf{O}$ is a by-product of muscle metabolism, and creatinine levels in the body are known to be a fairly reliable indicator of kidney function. The normal level of creatinine in the blood for adults is approximately 1.0 mg per deciliter (dL) of blood. If the density of blood is 1.025 g/mL, calculate the molality of a normal creatinine level in a 10.0-mL blood sample. What is the osmotic pressure of this solution at $\mathbf{25}\mathbf{.}{\mathbf{0}}^o\mathbf{C}$?

Short answer

$m=8.62\times {10}^{-5}mol/kg,\pi =2.15\times {10}^{-3}atm$

Step-by-step solution

Step 1:

In 10mL sample mass (blood) =10.25g.

$n\left(creatinine\right)=\frac{\text{mass}\left(creatinine\right)}{Molarmass\left(creatinine\right)}=\frac{1.0\times {10}^{-3}}{113.12g/mol}=8.8\times {10}^{-7}mol$

Step 2:

Calculate the osmotic pressure by the formula$\pi =MRT=\frac{n\left(creatinine\right)}{V_{solv}}RT$.

$\begin{array}{l}\pi =\frac{8.8\times {10}^{-7}}{0.01L}\times 0.08206Latm{K}^{-1}mol\times 298K\\ \pi =2.15\times {10}^{-3}atm\end{array}$