Chapter 2, Problem 37QAP

A cube of sodium has length 1.25 in. How many atoms are in that cube? (Note: $d_{\text{Na}}=0.968$ g/cm³.)

Interpretation Introduction

Interpretation:

To determine the number of atoms in given cube.

Concept introduction:

Relation between mass, volume and density:

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

Formula to calculate volume of cube is given as:

$\begin{array}{l}\text{Volume}\text{of}\text{cube}\text{=}{\text{a}}^{\text{3}}\\ \text{where}\text{'}\text{a'}\text{is}\text{length}\text{of}\text{side}\text{of}\text{cube}\text{.}\end{array}$

Relation between moles, mass of substance and number of atoms:

$\begin{array}{l}\text{moles}\text{=}\frac{\text{mass}\text{(g)}}{\text{molar}\text{mass}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{{\text{N}}_{\text{A}}}\\ \text{where,}\\ {\text{N}}_{\text{A}}\text{represents}\text{avagadro’s}\text{number}\text{equal}\text{to}\text{6}{\text{.023×10}}^{\text{23}}\text{atoms/mol}\end{array}$

Answer to Problem 37QAP

Number of atoms in given cube of sodium= $\text{8}{\text{.12×10}}^{\text{23}}\text{atoms}$

Explanation of Solution

Here we have:

Length of side of cube = 1.25 in

Since, 1 in = 2.54 cm

Therefore,

$\text{1}\text{.25}\text{inches}\text{=}\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}}\text{×1}\text{.25}\text{in}\text{=}\text{3}\text{.175}\text{cm}$

Formula to calculate volume of cube is given as:

$\begin{array}{l}\text{Volume}\text{of}\text{cube}\text{=}{\text{a}}^{\text{3}}\\ \text{where}\text{'}\text{a'}\text{is}\text{length}\text{of}\text{side}\text{of}\text{cube}\text{.}\end{array}$

Plugging values in above formula to get the volume of cube as:

$\begin{array}{l}\text{Volume}\text{of}\text{cube}\text{=}{\text{(3}}^{\text{.175)}}\\ \text{=}\text{32}\text{.01}{\text{cm}}^{\text{3}}\end{array}$

According to question, density of sodium = 0.968 g/cm³

Relation between mass, volume and density:

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

Plugging values in above formula to get the mass of sodium in 32.01 cm³volume as:

$\begin{array}{l}\text{0}\text{.968}{\text{g/cm}}^{\text{3}}\text{=}\frac{\text{Mass}}{\text{32}\text{.01}{\text{cm}}^{\text{3}}}\\ \text{Mass}\text{=}\text{0}\text{.968}\frac{\text{g}}{{\text{cm}}^{\text{3}}}\text{×32}\text{.01}{\text{cm}}^{\text{3}}\\ \Rightarrow \text{Mass}\text{=}\text{30}\text{.99}\text{g}\end{array}$

As we know:

$\text{moles}\text{=}\frac{\text{mass}\text{(g)}}{\text{molar}\text{mass}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{{\text{N}}_{\text{A}}}$

Therefore,

$\frac{\text{mass}\text{(g)}}{\text{molar}\text{mass}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{{\text{N}}_{\text{A}}}$

Here we have:

Mass of sodium = 30.99 g

Molar mass of sodium = 23.0 g/mol

Plugging values to get the number of atoms as:

$\begin{array}{l}\frac{\text{mass}\text{(g)}}{\text{molar}\text{mass}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{{\text{N}}_{\text{A}}}\\ \frac{\text{30}\text{.99}\text{(g)}}{\text{23}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{\text{6}{\text{.023×10}}^{\text{23}}}\\ \Rightarrow \text{number of}\text{atoms}\text{=}\frac{\text{30}\text{.99×6}{\text{.023×10}}^{\text{23}}}{\text{23}}\\ \Rightarrow \text{number}\text{of}\text{atoms}\text{=}\text{8}{\text{.12×10}}^{\text{23}}\text{atoms}\end{array}$

Conclusion

Number of atoms in given cube of sodium= $\text{8}{\text{.12×10}}^{\text{23}}\text{atoms}$