Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 30

The mass of \(1 \mathrm{mol}\) of \(^{13} \mathrm{C}\) (carbon- 13 ) is $13.003 \mathrm{g}$. (a) What is the mass in \(u\) of one \(^{13} \mathrm{C}\) atom? (b) What is the mass in kilograms of one \(^{13} \mathrm{C}\) atom?

Short Answer

Expert verified
Answer: The mass of one \(^{13} \mathrm{C}\) atom is \(13.003 \ \text{u}\) in atomic mass units and \(2.1611 \times 10^{-26} \ \text{kg}\) in kilograms.

Step by step solution

01

(Find the mass in atomic mass units)

Using the given mass of 1 mol of \(^{13} \mathrm{C}\) atoms which is \(13.003 \mathrm{g}\), we will first find the mass of one \(^{13} \mathrm{C}\) atom in atomic mass units (u). Since there are Avogadro's number (\(6.022 \times 10^{23}\) atoms) in 1 mol, we can find the mass of one \(^{13} \mathrm{C}\) atom in atomic mass units by dividing the molar mass by Avogadro's number: $$ \text{Mass in u} = \frac{13.003 \mathrm{g/mol}}{6.022 \times 10^{23} \text{atoms/mol}} $$
02

(Calculate the mass in atomic mass units)

Now we will plug in the numbers and multiply to get our result: $$ \text{Mass in u} = \frac{13.003 \text{g/mol}}{6.022 \times 10^{23} \text{atoms/mol}} \times \frac{1 \ \text{mol}}{6.022 \times 10^{23} \ \text{atoms}} = 13.003 \ \text{u} $$ So the mass of one \(^{13} \mathrm{C}\) atom in atomic mass units is \(13.003 \ \text{u}\).
03

(Find the mass in kilograms)

Next, we will find the mass of one \(^{13} \mathrm{C}\) atom in kilograms. To do this, we will use the conversion factor between grams and atomic mass units, which is: $$ 1 \ \text{u} = 1.6605 \times 10^{-27} \ \text{kg} $$ We can multiply the mass in atomic mass units by this conversion factor to find the mass in kilograms: $$ \text{Mass in kg} = 13.003 \ \text{u} \times \frac{1.6605 \times 10^{-27} \ \text{kg}}{1 \ \text{u}} $$
04

(Calculate the mass in kilograms)

Now we will plug in the numbers and multiply to get our result: $$ \text{Mass in kg} = 13.003 \ \text{u} \times \frac{1.6605 \times 10^{-27} \ \text{kg}}{1 \ \text{u}} = 2.1611 \times 10^{-26} \ \text{kg} $$ So the mass of one \(^{13} \mathrm{C}\) atom in kilograms is \(2.1611 \times 10^{-26} \ \text{kg}\). In conclusion, the mass of one \(^{13} \mathrm{C}\) atom is \(13.003 \ \text{u}\) in atomic mass units and \(2.1611 \times 10^{-26} \ \text{kg}\) in kilograms.

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

Given that our body temperature is \(98.6^{\circ} \mathrm{F},\) (a) what is the average kinetic energy of the molecules in the air in our lungs? (b) If our temperature has increased to \(100.0^{\circ} \mathrm{F},\) by what percentage has the kinetic energy of the molecules increased?
A cylinder in a car engine takes \(V_{i}=4.50 \times 10^{-2} \mathrm{m}^{3}\) of air into the chamber at \(30^{\circ} \mathrm{C}\) and at atmospheric pressure. The piston then compresses the air to one-ninth of the original volume \(\left(0.111 \mathrm{V}_{\mathrm{i}}\right)\) and to 20.0 times the original pressure \(\left(20.0 P_{\mathrm{i}}\right) .\) What is the new temperature of the air?
About how long will it take a perfume molecule to diffuse a distance of $5.00 \mathrm{m}\( in one direction in a room if the diffusion constant is \)1.00 \times 10^{-5} \mathrm{m}^{2} / \mathrm{s} ?$ Assume that the air is perfectly still-there are no air currents.
The temperature at which liquid nitrogen boils (at atmospheric pressure) is \(77 \mathrm{K}\). Express this temperature in (a) \(^{\circ} \mathrm{C}\) and (b) \(^{\circ} \mathrm{F}\).
What is the kinetic energy per unit volume in an ideal gas at (a) \(P=1.00\) atm and (b) \(P=300.0\) atm?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Oscillations

Read Explanation

Modern Physics

Read Explanation

Engineering Physics

Read Explanation

Electrostatics

Read Explanation

Electricity

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