`1 g` charcoal is placed in `100 mL` of `0.5 M CH_(3)COOH` to form an adsorbed mono-layer of acetic acid molecule and thereby the molarity of `CH_(3)COOH` reduces to `0.49`. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface are of charocal `=3.01xx10^(2)m^(2)//g`.

Options

A. `5 xx 10^(-19) m^(2)`
B. `5 xx 10^(20) m^(2)`
C. `5 xx 10^(-21) m^(2)`
D. `5 xx 10^(-23) m^(2)`

Answer

A

Explanation

Initial moles of acetic acid `= (0.5 xx 100)/(1000) = 0.05`
Moles of acetic acid left `= (0.49 xx 100)/(1000) = 0.049`
Moles of acetic acid absorbed `0.05 - 0.049 = 0.01` mol in a litre
Molecules `6.022 xx 10^(23) xx 0.001 = 6.022 xx 10^(2)` in `100 ml`
Area occupied by single molecule of acetic acid `= ("Total area")/("no. of molecules") = (3.01 xx 10^(2))/(6.022 xx 10^(20)) = 5 xx 10^(-19) m^(2)`