Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 81

An aqueous solution of 10.00 \(\mathrm{g}\) of catalase, an enzyme found in the liver, has a volume of 1.00 \(\mathrm{L}\) at \(27^{\circ} \mathrm{C}\) . The solution's osmotic pressure at \(27^{\circ} \mathrm{C}\) is found to be 0.745 torr. Calculate the molar mass of catalase.

Short Answer

Expert verified
The molar mass of catalase is approximately \(2.41 \times 10^5 \mathrm{g/mol}\).

Step by step solution

01

Convert the temperature to Kelvin and pressure to atm

First, we need to convert the given temperature from Celsius to Kelvin: \( T (K) = T ( °C) + 273.15 \) Next, we need to convert the osmotic pressure from torr to atm: 1 atm = 760 torr
02

Calculate the molar concentration of catalase

Equating the osmotic pressure formula, we get: Osmotic Pressure = (Molar Concentration) × (R) × (Temperature in K) Rearranging to find Molar Concentration: Molar Concentration = (Osmotic Pressure) / (R × Temperature in K) Now, substitute the given values and the converted values from Step 1 into the equation and solve for the molar concentration: R (ideal gas constant) = 0.0821 L atm / (K mol)
03

Find the number of moles of catalase

Using the calculated molar concentration and the volume of the solution, we can find the number of moles of catalase: Number of moles of catalase = (Molar Concentration) × (Volume of solution in L)
04

Calculate the molar mass of catalase

Finally, we can use the formula for molar mass to find the molar mass of catalase: Molar Mass = (mass of solute) / (number of moles of solute) Now, substitute the values obtained from Steps 2 and 3 into the formula and find the molar mass of catalase.

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

A 0.500 -g sample of a compound is dissolved in enough water to form 100.0 mL of solution. This solution has an osmotic pressure of 2.50 atm at $25^{\circ} \mathrm{C}$ . If each molecule of the solute dissociates into two particles (in this solvent), what is the molar mass of this solute?

Consider a beaker of salt water sitting open in a room. Over time, does the vapor pressure increase, decrease, or stay the same? Explain.

Thyroxine, an important hormone that controls the rate of metabolism in the body, can be isolated from the thyroid gland. When 0.455 g thyroxine is dissolved in 10.0 g benzene, the freezing point of the solution is depressed by \(0.300^{\circ} \mathrm{C}\) . What is the molar mass of thyroxine? See Table \(11.5 .\)

You have a solution of two volatile liquids, A and B (assume ideal behavior). Pure liquid A has a vapor pressure of 350.0 torr and pure liquid B has a vapor pressure of 100.0 torr at the temperature of the solution. The vapor at equilibrium above the solution has double the mole fraction of substance A that the solution does. What is the mole fraction of liquid A in the solution?

If a solution shows positive deviations from Raoult’s law, would you expect the solution to have a higher or lower boiling point than if it were ideal? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

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