Chapter 13, Problem 1P

How does the number of atoms in a 27.5-gram gold ring compare to the number in a silver ring of the same mass?

To determine

The ratio of the number of atoms in a gold ring to that in a silver ring of the given mass

Answer to Problem 1P

Solution:

The ratio of the number of atoms in a gold ring of 27.5 g to the number of atoms in a silver ring of the same mass is 0.547.

Explanation of Solution

Given:

Mass of the gold ring, $m_g=27.5g$

Mass of the silver ring, $m_s=27.5g$

Formula used:

The number of moles of a given specimen of substance is given by,

$n=\frac{m}{M}$, where m is the mass of the specimen and M is the mass of 1 mole of the substance

Calculation:

Since gold and silver both have monatomic molecules, the ratio of number of atoms = the ratio of number of moles

The number of moles of a given specimen of substance is given by,

$n=\frac{m}{M}$, where m is the mass of the specimen and M is the mass of 1 mole of the substance

For gold, $M_g=196.97\frac{g}{mol}$

For silver, $M_s=107.8\frac{g}{mol}$

Therefore, the required ratio is

$\frac{n_g}{n_s}=\frac{(\frac{m_g}{M_g})}{(\frac{m_s}{M_s})}$

= $\frac{(\frac{27.5g}{196.97\frac{g}{mol}})}{(\frac{27.5g}{107.8\frac{g}{mol}})}$

= $\frac{107.8\frac{g}{mol}}{196.97\frac{g}{mol}}$

= 0.547