One gram of charcoal adsorbs 100 mL of 0.5 `M CH_(3)COOH` to form a mono-layer and thereby the molarity of acetic acid is reduced to 0.49 M. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface acid of charcoal `=3.01xx10^(2) m^(2)//gm`

Options

A. `0.5 xx 10^(-19) m^(2)`
B. `4xx 10^(-20) m^(2)`
C. `0.5 xx 10^(-10) m^(2)`
D. `5 xx10^(-19) m^(2)`

Answer

D

Explanation

Amount of CH,COOII adsorbed on 1y of charcoal`=(0.5-0.49moL^(-1))xx(1000mL)/(1000mL//L)=1xx10^(-3)mol`
So, No. of CH,COOH molecules in `1xx10^(-3) mol` of acetic acid
`=6.02 10^(23)mol^(-1) xx10^(-3)mol =6.02 xx 10^(20)`
Area occupied by one CH,COOH molecule
`=(93.01xx10^(2)m^(2))/(6.02xx10^(20))=5xx10^(-19)m^(2)`
Thus the surface area of the charcoal covered by each `CH_(3)COOH` molecule is `5 xx 10^(-9)m^(2)`