91Ó°ÊÓ

The given data is listed below,

The distance between uneven surfaces of book and table is, $t=0.1mm×\frac{1m}{1000mm}=0.0001m$

Assume the other data as follows,

The number of moles of air between two surfaces depends on the space between them and the volume of objects.

The moles of air can be obtained by dividing the volume of the object by the molar volume of an ideal gas.

Write the expression for the volume of air between the surfaces.

$V=I×b×t$

Here, I is the length of the book, b is the breadth of the book, and t is the distance between uneven surfaces of book and table.

Substitute all the values in the above expression.

$\begin{array}{rcl}V& =& 20cm×30cm×0.0001m\\ & =& \left(20cm×\frac{1m}{100cm}\right)×\left(30cm×\frac{1m}{100cm}\right)×0.0001m\\ & =& 0.2m×0.3m×0.0001m\\ & =& 6×{10}^{-5}{m}^3\end{array}$

The expression for the moles of air is given as,

$n=\frac{V}{V_m}$

Here, V volume of air between the surfaces, and V_m is the molar volume of air.

Substitute all the values in the above expression.

$\begin{array}{rcl}n& =& \frac{6×{10}^{-5}{m}^3}{0.224m^3/mol}\\ & =& 2.678×{10}^{-5}mol\end{array}$

The expression for the number of molecules between textbook and table is give as,

N = nN_a

Here, n is the number of moles of air, and N_a is Avogadro’s number with the value $(6.022×{10}^{23}mo{l}^{-1})$.

Substitute all the values in the above expression.

$\begin{array}{rcl}N& =& 2.678×{10}^{-5}mol×(6.022×{10}^{23}mo{l}^{-1})\\ & =& 1.61×{10}^{19}\end{array}$

Thus, the number of molecules between textbook and table is $1.61×{10}^{19}$.