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2.0 g of charcoal is placed in 100 mL of...

`2.0 g` of charcoal is placed in `100 mL` of `0.05 M CH_(3)COOH` to form an adsorbed mono-acidic layer of acetic acid molecules and thereby the molarity of `CH_(3)COOH` reduces to `0.49`. The surface area of charcoal is `3xx10^(2) m^(2)g^(-1)`. The surface area of charcoal is adsorbed by each molecule of acetic acid is
a. `1.0xx10^(-18)m^(2)` b. `1.0xx10^(-19)m^(2)`
c. `1.0xx10^(13)m^(2)` d. `1.0xx10^(-22)m`

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To solve the problem, we will follow these steps: ### Step 1: Calculate the decrease in molarity of acetic acid The initial molarity of acetic acid is given as `0.05 M`, and the final molarity after adsorption is `0.49 M`. The decrease in molarity can be calculated as follows: \[ \Delta M = M_{\text{initial}} - M_{\text{final}} = 0.05 - 0.49 = 0.01 \, \text{M} \] ...
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