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 \(25.0^{\circ} \mathrm{C} ?\)

The problem states that the normal level of creatinine in the blood is approximately 1.0 mg per dL of blood. To find the mass of creatinine in the 10.0 mL blood sample, first, convert mL to dL: \(1~\text{dL} = 100~\text{mL}\) So, \(10.0~\text{mL} = 0.1~\text{dL}\) Now multiply the creatinine concentration with the volume of the blood sample: Mass of creatinine = \((0.1~\text{dL}) \times (1.0~\text{mg}/\text{dL}) = 0.1~\text{mg}\) Now, convert the mass from mg to g: \(0.1~\text{mg} = 0.1 \times 10^{-3}~\text{g} = 1.0 \times 10^{-4}~\text{g}\)

We are given the density of blood as \(1.025~\text{g/mL}\). To find the mass of a 10.0 mL blood sample, use the formula: Mass = Density × Volume \(m_\text{blood} = (1.025~\text{g/mL}) \times (10.0~\text{mL}) = 10.25~\text{g}\)

Molality is defined as the moles of solute per kilogram of solvent. First, calculate the moles of creatinine: The molar mass of creatinine, \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~N}_{3} \mathrm{O}\), is: Molar mass = 4(12.01) + 7(1.01) + 3(14.01) + 16.00 = 48.04 + 7.07 + 42.03 + 16.00 = 113.14 g/mol Therefore, the moles of creatinine = mass of creatinine / molar mass: moles = \( \frac{1.0 \times 10^{-4}~\text{g}}{113.14~\text{g/mol}} = 8.84 \times 10^{-7}~\text{mol}\) Now calculate the molality of creatinine: Molality = \( \frac{8.84 \times 10^{-7}~\text{mol}}{0.01025~\text{kg}} = 8.62 \times 10^{-5}~\text{mol/kg}\)

We are given the temperature as \(25.0^{\circ} \mathrm{C}\), which in kelvin scale is: \(T = 25.0 + 273.15 = 298.15~\text{K}\) Now, use the formula for osmotic pressure: Osmotic pressure = molality × R × T Here, R is the gas constant, which is equal to \(8.314~\text{J/mol·K}\). Osmotic pressure = \( (8.62 \times 10^{-5}~\text{mol/kg}) \times (8.314~\text{J/mol·K}) \times (298.15~\text{K}) = 212.68 \times 10^{-5}~\text{J/kg} \) Since we usually report osmotic pressure in atmospheres (atm), we'll convert from J/kg to atm: \(1~\text{atm} = 101,325~\text{J/m³}\) Because \(1~\text{m³} = 1,000~\text{kg}\), we have: \(1~\text{atm} = 101,325 / 1,000~\text{J/kg} = 101.325~\text{J/kg}\) Therefore, the osmotic pressure of the solution is: \(212.68 \times 10^{-5}~\text{J/kg} × \frac{1~\text{atm}}{101.325~\text{J/kg}} = 2.10 \times 10^{-6}~\text{atm}\)