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Chapter 10: Problem 130

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 -mL blood sample. What is the osmotic pressure of this solution at \(25.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The molality of creatinine in the 10.0 mL blood sample is approximately \(8.61 \times 10^{-5} m\), and the osmotic pressure of this solution at 25.0 °C is about 0.00221 atm.

Step by step solution

01

Calculate the mass of creatinine in the blood sample

Given that the normal level of creatinine is 1.0 mg/dL, we should first find the mass of creatinine in the 10.0 mL blood sample. Since there are 100 mL in 1 dL, we can calculate the mass of creatinine as follows: \(mass \, of \, creatinine = \frac{1.0 \, mg}{1 \, dL} \times \frac{10.0 \, mL}{100 \, mL}\) \(mass \, of \, creatinine = 0.1 \, mg\) To convert it to grams, we simply multiply by the conversion factor: \(mass \, of \, creatinine (grams) = 0.1 \, mg \times \frac{1 \, g}{1000 \, mg} = 0.0001 \, g\)
02

Calculate moles of creatinine

Now we need to convert grams of creatinine to moles. We can use the formula: \( moles \, of \, creatinine = \frac{mass \, of \, creatinine}{molar \, mass \, of \, creatinine} \) Where molar mass of creatinine = 4(12.01 g/mol) + 7(1.01 g/mol) + 3(14.01 g/mol) + 1(16.00 g/mol) = 113.15 g/mol \( moles \, of \, creatinine = \frac{0.0001 \, g}{113.15 \, g/mol} \approx 8.83 \times 10^{-7} \, mol\)
03

Calculate mass of solvent (blood)

Next, we should find the mass of blood in the 10.0 mL sample. Using the given density: \(mass \, of \, blood (g) = volume \, of \, blood \times density \, of \, blood\) \(mass \, of \, blood (g) = 10.0 \, mL \times \frac{1.025 \, g}{1 \, mL} = 10.25 \, g\)
04

Calculate molality of creatinine

Now, we can calculate the molality of creatinine using the formula: \(molality (m) = \frac{moles \, of \, solute}{mass \, of \, solvent (kg)}\) \(molality (m) = \frac{8.83 \times 10^{-7} \, mol}{0.01025 \, kg} \approx 8.61 \times 10^{-5} \, m\)
05

Calculate the osmotic pressure

Finally, we need to find the osmotic pressure of this solution at 25.0 °C. We'll use the formula: \(osmotic \, pressure = molality \times R \times T\) Where R is the gas constant (0.0821 L atm/mol K) and T is the temperature in Kelvin (25.0 + 273.15 = 298.15 K). \(osmotic \, pressure = (8.61 \times 10^{-5} \,m)\times(0.0821\, \frac{L\, atm}{mol\, K}) \times (298.15\, K)\) \(osmotic \, pressure \approx 0.00221 atm\) So, the molality of creatinine in the blood sample is approximately \(8.61 \times 10^{-5} m\), and the osmotic pressure of this solution at 25.0 °C is about 0.00221 atm.

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