Question

Finely divided catalyst has greater surface area and has greater catalytic activity than the compact solid. If a total surface area of $6291456c{m}^2$ is required for adsorption in a catalyzed gaseous reaction, then how many splits should be made to a cube of exactly $1cm$ in length to achieve required surface area? (Given: One split of a cube gives eight cubes of same size)

Answer 1

$\text{Finding the edge length on}{1}^{st}\text{split}$



Let edge length of cube $=a$

$1^{st}$ split gives 8 cubes.

Edge length on $1^{st}$ split $=\frac{a}{2}cm$

$\text{Finding the edge length on}{2}^{nd}\text{split}$

$2^{nd}$ split gives $8^2$cubes.

Edge length on $2^{nd}$ split $=\frac{a}{2^2}cm$

Finding the edge length on $n^{th}$ split

$n=$ number of splits

$n^{th}$ split gives $8^n$ cubes.

Edge length on $n^{th}$ split $=\frac{a}{2^n}cm$

$\text{Finding the value of number of splits}$

Surface area of a cube $=6\times (side{)}^2$

Edge length on $n^{th}$ split $=\frac{a}{2^n}cm$

Total surface area of cube $=6\times {\left(\frac{a}{2^n}cm\right)}^2$

Total surface area of $8^n$ number of cubes:
$=6\times 8^n\times {\left(\frac{a}{2^n}cm\right)}^2$

Given surface area of cube having edge length of $1cm$ is $6291456c{m}^2.$

So,

$6\times 8^n\times {\left(\frac{1}{2^n}cm\right)}^2=6291456c{m}^2$

$\Rightarrow 8^n\times \left(\frac{1}{4^n}\right)=1048576$

$\Rightarrow 2^n=1048576$

$\Rightarrow 2^n=2^{20}$

$\Rightarrow n=20splits$

So, option (c) is the correct answer.