Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 140

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 \((\mathrm{dL})\) 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 approximately \(8.61 \times 10^{-5}\) mol/kg. The osmotic pressure of this solution at 25.0°C is approximately 0.0215 Pa.

Step by step solution

01

Convert the given information into the required units

We start by converting 10.0 mL blood sample to mass and 1.0 mg/dL creatinine level to moles. Density of blood = 1.025 g/mL Mass of blood = Density × Volume = 1.025 g/mL × 10.0 mL = 10.25 g 1 dL = 100 mL Creatinine level = 1.0 mg/dL = 0.01 mg/mL Mass of creatinine in the blood sample = 0.01 mg/mL × 10.0 mL = 0.1 mg Given that the molar mass of creatinine (C\(_4\)H\(_7\)N\(_3\)O) is roughly M = 113.12 g/mol. Now we will convert the mass of creatinine in our sample from mg to g: Mass of creatinine = 0.1 mg × (1 g/1000 mg) = 0.0001 g Now we will convert grams of creatinine to moles using its molar mass: Moles of creatinine = Mass / Molar mass = 0.0001 g / 113.12 g/mol = 8.83 × 10\(^{-7}\) mol
02

Calculate molality of the solution

Now that we have the moles of creatinine in the blood sample, we can calculate the molality. The formula for molality is: Molality = Moles of solute / Mass of solvent (in kg) In our case, the solute is creatinine and the solvent is blood. Molality = 8.83 × 10\(^{-7}\) mol / 0.01025 kg = 8.61 × 10\(^{-5}\) mol/kg
03

Calculate the osmotic pressure of the solution

Now we will calculate the osmotic pressure of the solution using the following formula: Osmotic pressure (π) = n × R × T where n = molality of the solution = 8.61 × 10\(^{-5}\) mol/kg R = gas constant = 8.314 J/mol·K (since we are using SI units) T = temperature in Kelvin = 273.15 + 25.0 = 298.15 K We can now plug in these values into the osmotic pressure formula: Osmotic pressure (π) = 8.61 × 10\(^{-5}\) mol/kg × 8.314 J/mol·K × 298.15 K = 0.0215 Pa The osmotic pressure of this solution at 25.0°C is approximately 0.0215 Pa.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is $2.63^{\circ} \mathrm{C}\left(K_{\mathrm{f}}\right.\( for camphor is \)40 .^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}$ ). Calculate the molality of the solution and the molar mass of reserpine.

The normal boiling point of methanol is \(64.7^{\circ} \mathrm{C} .\) A solution containing a nonvolatile solute dissolved in methanol has a vapor pressure of 556 torr at \(64.7^{\circ} \mathrm{C} .\) What is the mole fraction of methanol in this solution?

Which of the following statements is(are) true? Correct the false statements. a. The vapor pressure of a solution is directly related to the mole fraction of solute. b. When a solute is added to water, the water in solution has a lower vapor pressure than that of pure ice at \(0^{\circ} \mathrm{C} .\) c. Colligative properties depend only on the identity of the solute and not on the number of solute particles present. d. When sugar is added to water, the boiling point of the solution increases above \(100^{\circ} \mathrm{C}\) because sugar has a higher boiling point than water.

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57\(\% \mathrm{C}\) and 5.30\(\%\) H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of $-5.20^{\circ} \mathrm{C}\( is recorded for a solution made by dissolving 10.56 \)\mathrm{g}$ of the compound in 25.0 \(\mathrm{g}\) water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Would benzoic acid be more or less soluble in a \(0.1-M \mathrm{NaOH}\) solution than it is in water?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free