Chapter 13, Problem 70A

Interpretation Introduction

Interpretation:

The element X in the given sample ${\text{X}}_{\text{4}}{\text{H}}_{\text{10}}$ is to be identified.

Concept introduction:

The ideal gases are those gases which obey the ideal gas equation and the ideal gas equation is written as

  $\text{PV}\text{= nRT}$

Answer to Problem 70A

The element X is Boron with atomic number 11.

Explanation of Solution

Given information:

Volume = $\text{30}{\text{.0cm}}^{\text{3}}$

Volume in liter = $\text{30}{\text{.0×10}}^{\text{-3}}\text{L}$

Temperature = $\text{20°C}\text{=}\text{20}\text{+}\text{273=}\text{293K}$

Pressure = $\text{801}\text{mmHg}$

Pressure in atmospheric pressure ₌$\frac{\text{801}}{\text{760}}\text{=}\text{1}\text{.05}\text{atm}$

R, gas constant = $\text{0}\text{.0821}$

Calculation:

According to ideal gas equation,

  $\begin{array}{l}\text{n=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{1}{\text{.054×30×10}}^{\text{-3}}}{\text{0}\text{.0821×293}}\\ \text{=}1.314{\text{×10}}^{\text{-3}}\text{moles}\end{array}$

The given sample containing $1.314{\text{×10}}^{\text{-3}}\text{moles}$ = $\text{0}\text{.0712g}$

And, 1 mole = $\frac{\text{0}\text{.0712}}{\text{1}{\text{.31×10}}^{\text{-3}}}\text{=}\text{54}\text{.3g}$

Total molecular weight of the sample is $\text{54}\text{.3g}$

In the given sample, hydrogen contains 10gm and remaining weight is the molecular weight of the element X

So, molecular weight of X = $\text{44}\text{.3g}$

Now,

  $\begin{array}{l}\text{4X}\text{=}\text{44}\text{g}\\ \text{X=}\frac{\text{44}}{\text{4}}\\ \text{X}\text{=}\text{11}\end{array}$

The element with atomic number 11 is Boron.

So, the compound is ${\text{B}}_{\text{4}}{\text{H}}_{\text{10}}$

Conclusion

The element X in ${\text{X}}_{\text{4}}{\text{H}}_{\text{10}}$ is Boron.