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Q. Calculate the surface area of a catalyst (in terms of $10^{3}$ cm$^{2}$) that adsorbs $10^{3}$ cm$^{3}$ of nitrogen reduced to $STP$ per gram in order to form the monolayer. The effective area occupied by $N_{2}$ molecule on the surface is $1.62 \times 10^{-15}$ cm$^{2}$

Surface Chemistry

Solution:

No. of $N_{2}$ molecules$= \frac{10^{3} \times6.023 \times10^{23}}{22400} = 2.69 \times10^{22}$
Total area covered by $N_{2} = 2.69 \times10^{22} \times1.62 \times10^{-15}$
$ = 435 \times10^{5}$ cm$^{2}$
$ = 4.35 \times10^{3}$ cm$^{2}$