The question is:
If $2.0 \cdot 10^{-4}$ moles of dye in $\pu{50 mL}$ of solution is consumed in 188 seconds, what is the average rate of consumption of dye in $\pu{mol L-1 s-1}$?
I am not sure if I have done this correctly. I use the rate of reaction formula which is $\text{Rate} = -\frac{\mathrm{d[A]}}{\mathrm dt}$ There is a negative sign because the dye is being consumed and it is disappearing.
Given:
[dye] = $2.0 \cdot 10^{-4}$ moles in $\pu{50 mL}$
volume of solution = $\pu{50 mL}$ = $\pu{0.05 L}$
$t = \pu{188 s}$
Since we have $2.0 \cdot 10^{-4}$ moles of dye per $\pu{50 mL}$ of solution, that means in $\pu{1 L}$ of solution, we get $2*(2.0 \cdot 10^{-4})$ moles of dye = $4.0 \cdot 10^{-4}$ moles
\begin{align} \text{Rate of consumption/disappearance} &= \frac{-4.0 \cdot 10^{-4} \, \pu{mol}}{\pu{188 s}} \\ &= -2.1 \cdot 10^{-6} \pu{mol L-1 s-1} \end{align}
Is my answer correct? Please point out my mistakes if I did it incorrectly.