Answer : The time taken for the reaction is, 28 s.
Explanation :
Expression for rate law for first order kinetics is given by :
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 0.0632
t = time taken for the process = ?
= initial amount or concentration of the reactant = 1.28 M
= amount or concentration left time 't' = 
Now put all the given values in above equation, we get:


Therefore, the time taken for the reaction is, 28 s.
<span>moles glucose = 19 g / 180 g/mol= 0.105
M = 0.105 / 0.100 L = 1.05
moles in 20.0 mL = 1.05 M x 0.0200 L = 0.0216
New concentration = 0.0216 /0.500 L = 0.0432 M
moles in 100 mL = 0.100 L x 0.0432 = 0.00432
mass = 0.00432 x 180 g/mol= 0.778 g</span>
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