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Chapter 3: Problem 30

(a) What is the mass, in grams, of one mole of \({ }^{79} \mathrm{Br}\) ? (b) How many bromine atoms are present in one mole of \({ }^{79} \mathrm{Br}\) ?

Short Answer

Expert verified
(a) The mass of one mole of bromine-79 is 79 g. (b) There are \(6.022 \times 10^{23}\) bromine atoms in one mole of bromine-79.

Step by step solution

01

(a) Determine the mass of one mole of bromine-79

To determine the mass of one mole of bromine-79, we need to know the atomic mass of bromine-79. The atomic mass of an isotope is approximately equal to its mass number, and the given isotope is \({ }^{79} \mathrm{Br}\) , which has a mass number of 79. Therefore, the atomic mass of bromine-79 is 79 atomic mass units (amu). To convert this to grams, we will use the definition of a mole, which states that one mole of a substance contains Avogadro's number (\(6.022 \times 10^{23}\)) of particles and has a mass in grams equal to the atomic mass in amu. So: Mass of one mole of bromine-79 = Atomic mass in amu × 1 g/mol = 79 amu × 1 g/mol
02

(a) Calculate the mass in grams

Now that we have the formula, we can calculate the mass of one mole of bromine-79: Mass of one mole of bromine-79 = 79 g/mol This means that one mole of bromine-79 has a mass of 79 grams.
03

(b) Determine the number of bromine atoms in one mole of bromine-79

We know that one mole of a substance contains Avogadro's number of particles. So, in our case, one mole of bromine-79 contains Avogadro's number of bromine atoms. This can be calculated as: Number of bromine atoms in one mole of bromine-79 = Avogadro's number = \(6.022 \times 10^{23}\) atoms This means that one mole of bromine-79 contains \(6.022 \times 10^{23}\) bromine atoms.

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Most popular questions from this chapter

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