Chapter 20, Problem 68E

Interpretation Introduction

Interpretation:

Whether the radioactivity of $\text{US}$ or ${\text{US}}_{\text{2}}$ is more is to be predicted. If $\text{0}\text{.5}\text{moles}$ of each compound is prepared, the compound which is more radioactive is to be stated. The explanation for both the answers is to be stated.

Concept introduction:

Half-life is defined as the time required for the number of nuclides to reach half of the original value. Half-life is defined as the time required for the number of nuclides to reach half of the original value. The remaining mass of a sample is calculated by the formula shown below.

$\text{R}=\text{S}\times {\left(\text{1}/\text{2}\right)}^{\text{n}}$

Where,

• $\text{n}$ is the number of half-lives.

• $\text{R}$ is the remaining amount of a sample after $\text{n}$ half-lives.

• $\text{S}$ is the initial amount of a sample.

Answer to Problem 68E

The compound $\text{US}$ will exhibit more amount of radioactivity.

The radioactivity of $0.5$ mole of $\text{US}$ is same as the radioactivity of $0.5$ mole of ${\text{US}}_2$.

Explanation of Solution

The amount of uranium in $\text{50}\text{g}$ of $\text{US}$ is calculated by the formula shown below.

$\text{Ammount}\text{of}\text{U}\text{in}\text{US}=5\text{0}\text{g}\times \frac{\text{Molar}{\text{mass}}_{\text{U}}}{\text{Molar}{\text{mass}}_{\text{US}}}$...(1)

The value of molar mass of uranium and $\text{US}$ is $\text{238}\text{g}/\text{mol}$ and $\text{270}\text{g}/\text{mol}$ respectively.

Substitute the value of molar mass of uranium and $\text{US}$ in equation (1).

$\begin{array}{rll}\text{Ammount}\text{of}\text{U}\text{in}\text{US}& =& 5\text{0}\text{g}\times \frac{\text{Molar}{\text{mass}}_{\text{U}}}{\text{Molar}{\text{mass}}_{\text{US}}}\\ & =& 5\text{0}\text{g}\times \frac{\text{238}\text{g}/\text{mol}}{\text{270}\text{g}/\text{mol}}\\ & =& 5\text{0}\times 0.881\text{g}\\ & =& 44.05\text{g}\end{array}$

The amount of uranium in $\text{50}\text{g}$ of ${\text{US}}_2$ is calculated by the formula shown below.

$\text{Ammount}\text{of}\text{U}\text{in}{\text{US}}_2=5\text{0}\text{g}\times \frac{\text{Molar}{\text{mass}}_{\text{U}}}{\text{Molar}{\text{mass}}_{{\text{US}}_2}}$...(2)

The value of molar mass of uranium and ${\text{US}}_2$ is $\text{238}\text{g}/\text{mol}$ and $302\text{g}/\text{mol}$ respectively.

Substitute the value of molar mass of uranium and ${\text{US}}_2$ in equation (2).

$\begin{array}{rll}\text{Ammount}\text{of}\text{U}\text{in}{\text{US}}_2& =& 5\text{0}\text{g}\times \frac{\text{Molar}{\text{mass}}_{\text{U}}}{\text{Molar}{\text{mass}}_{\text{US}}}\\ & =& 5\text{0}\text{g}\times \frac{\text{238}\text{g}/\text{mol}}{302\text{g}/\text{mol}}\\ & =& 5\text{0}\times 0.788\text{g}\\ & =& 39.4\text{g}\end{array}$

The amount of uranium in $\text{50}\text{g}$ of $\text{US}$ is more than the amount present in $\text{50}\text{g}$ of ${\text{US}}_2$. Therefore, $\text{50}\text{g}$ of $\text{US}$ will exhibit more radioactivity.

The amount of any substance present in one mole is equal to the molar mass of that substance.

Therefore, the amount of uranium in one mole of $\text{US}$ is $\text{238}\text{g}$. The amount of uranium in one mole of ${\text{US}}_2$ is also $\text{238}\text{g}$.

Therefore the $\text{0}\text{.5}\text{mole}$ of $\text{US}$ and ${\text{US}}_2$ will have the same amount of uranium. Therefore, the radioactivity of $0.5$ mole of $\text{US}$ is same as the radioactivity of $0.5$ mole of ${\text{US}}_2$.

Conclusion

The compound $\text{US}$ will exhibit more amount of radioactivity.

The radioactivity of $0.5$ mole of $\text{US}$ is same as the radioactivity of $0.5$ mole of ${\text{US}}_2$.