1 g of charcoal adsorbs 100 mL of 0.5 M $C{H}_{3}COOH$ to form a monolayer and thereby the molarity of $C{H}_{3}COOH$reduces to 0.49. The surface area of charcoal adsorbed by each molecule of acetic acid (surface area of charcoal =$3.01\times {10}^2{m}^2/g)$   is ___ $\times {10}^{-19}{m}^2$

Millimole of acetic acid taken = 100 x 0.5 = 50
Millimole of acetic acid left = 100 x 0.49 = 49
Millimole of acetic acid adsorbed = 50 – 49 = 1
Molecules of acetic acid adsorbed    

$=1\times {10}^{-3}\times 6.023\times {10}^{23}=6.023\times {10}^{20}$      
Total area of 1 g charcoal covered by these molecules  $=3.01\times {10}^2{m}^2$
 $\therefore$Area covered by 1 molecule    $=\frac{3.01\times {10}^2}{6.023\times {10}^{20}}(\because unilayeradsorption)$

$=5\times {10}^{-19}{m}^2$