Chapter 9, Problem 1E

Interpretation Introduction

Interpretation:

The density of the recently discovered element $114$ to be estimated.

Concept introduction:

Density of the element is calculated by dividing the molar mass ( $\frac{g}{mol}$ ) by atomic (molar) volume ( $\frac{c{m}^3}{mol}$ ).

$Density(d)=\frac{MolarMass}{AtomicVolume}$

Answer to Problem 1E

The density of the recently discovered element $114$ is $16\frac{g}{c{m}^3}$.

Explanation of Solution

From periodic table and from the given data table, element having atomic number $114$ will be in the same group as $Pb$, an element of atomic number $82$ and atomic volume $18\frac{c{m}^3}{mol}$ and $Sn$, an element of atomic number $50$ and atomic volume $18\frac{c{m}^3}{mol}$. It’s noteworthy to mention that the atomic volumes of $Pb$ and $Sn$ are equal, may be due to lanthanide contraction. If there is also actinide contraction then also the atomic volume will be $18\frac{c{m}^3}{mol}$.

Thus, density of the element having atomic number $114$ is

$Density(d)=\frac{298\frac{g}{mol}}{18\frac{c{m}^3}{mol}}=16\frac{g}{c{m}^3}$

Conclusion

$16\frac{g}{c{m}^3}$ is the density of the recently discovered element $114$ .