Chapter 17, Problem 74A

To determine

To Compare:

The atomic masses of elements of A and B.

Answer to Problem 74A

The atomic mass of element A is $\frac{3}{2}\text{ or }1.5\text{times}$ atomic mass of element B.

Explanation of Solution

Given:

Gas A is diatomic while gas $\text{B}$ is monatomic.

Gas A has 3 times the mass of Gas B. Both have an equal volume at equal temperature and pressure.

Formula used:

As per the ideal gas equation, $PV=nRT$

Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature.

$M=nm$

Where, m is the mass, n is the number of moles and M is the molar mass.

Calculation:

As the pressure, volume, and temperature is the same for both the gases, the number of moles ( n ) would also be the same according to ideal gas law.

$\begin{array}{l}{n}_A=n_B\\ \Rightarrow \frac{m_A}{M_A}=\frac{m_B}{M_B}\\ \because m_A=3m_B\\ \Rightarrow M_A=3M_B\end{array}$

Gas A is diatomic; thus, it can be represented as ${\text{A}}_{\text{2}}$ and gas B is monoatomic, thus it can be represented as B. Atomic mass for A and B is related to molar mass as follows:

$M_A=2$ (Atomic mass of A )

$M_B=$ (Atomic mass of B )

Thus, 2(Atomic mass of A)=3 (Atomic mass of B)

The atomic mass of A $=\frac{3}{2}$ Atomic mass of B

Conclusion:

Therefore, the atomic mass of element A is 1.5 timesthe atomic mass of element B.