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Chapter 11: Problem 118

Creatinine, \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~N}_{3} \mathrm{O}\), 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 \(1.0 \mathrm{mg}\) per deciliter (dL) of blood. If the density of blood is \(1.025 \mathrm{~g} / \mathrm{mL}\), calculate the molality of a normal creatinine level in a \(10.0-\mathrm{mL}\) blood sample. What is the osmotic pressure of this solution at \(25.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The molality of a normal creatinine level in a 10.0 mL blood sample is \(0.0862 \: mol/kg\), and the osmotic pressure of this solution at \(25.0^{\circ} \mathrm{C}\) is \(2.12 \times 10^{-3} \: atm\).

Step by step solution

01

Convert mass of creatinine to moles

To begin, we convert the mass of creatinine to moles using its molar mass. The molar mass of creatinine is found by using the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{N}_{3} \mathrm{O}\): \(Molar\_mass\_of\_creatinine = 4 \times 12.01 + 7 \times 1.008 + 3 \times 14.01 + 16.00 = 113.12 \: g/mol\) From the given information, we know that the mass of creatinine in 1 dL (100 mL) of blood is 1.0 mg. Therefore, in a 10.0 mL blood sample, there would be 0.1 mg of creatinine. \(moles\_of\_creatinine = \frac{0.1 \: mg}{113.12 \: g/mol} = \frac{0.0001 \: g}{113.12 \: g/mol} = 8.84 \times 10^{-7} \: moles\)
02

Calculate the mass of solvent (blood) in the given blood sample

Now, we need to obtain the mass of blood in the 10.0 mL blood sample. We use the given density of blood (1.025 g/mL) to convert the volume of blood to mass. \(mass\_of\_blood = 10.0 \: mL \times \frac{1.025 \: g}{1 \: mL} = 10.25 \: g\)
03

Calculate the molality of creatinine

Now, we can calculate the molality of the creatinine in the blood sample. Molality is defined as the number of moles of solute per kilogram of solvent: \(molality = \frac{moles\_of\_creatinine}{mass\_of\_solvent (in \: kg)} = \frac{8.84 \times 10^{-7} \: moles}{0.01025 \: kg}\) \(molality = 0.0862 \: mol/kg\)
04

Calculate the osmotic pressure

Finally, we can calculate the osmotic pressure using the Van't Hoff equation: \(Π = i \times m \times R \times T\) Π = osmotic pressure (in atm) i = Van't Hoff factor (assumed to be 1 for non-electrolyte creatinine) m = molality (0.0862 mol/kg) R = gas constant (0.0821 L atm / K mol) T = temperature in Kelvin (25.0°C + 273.15 = 298.15 K) \(Π = (1) \times (0.0862 \: mol/kg) \times (0.0821 \: L \: atm / K \: mol) \times (298.15 \: K)\) \(Π = 2.12 \times 10^{-3} \: atm\) So, the osmotic pressure of the solution is \(2.12 \times 10^{-3} \: atm\).

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