Determination of the iodine value of a fatty oil: Does the concentration of Na2SO3 effect the iodine value?

I want to determine the iodine value of a fatty oil. For this I dissolve $$\ce{1.0 g}$$ of the oil in chloroform. Iodine bromide is then added and placed in a dark room for at least $$\pu{30 min}$$. After this, $$\ce{KI}$$ is added. Then, I am going to titrate with $$\pu{0.1 M}$$ $$\ce{Na2SO3}$$. Based on the added volume, I can calculate the iodine value of the unknown fatty oil.

Now I was wondering what the effect should be using $$\pu{0.02 M}$$ $$\ce{Na2SO3}$$ instead of $$\pu{0.1 M}$$? And is there a reason why the solution has to stay $$\pu{30 min}$$ in a dark room? I think there's no difference in outcome when it will be in a dark room, for example, say $$\pu{3 h}$$.

The reaction of $$\ce{I2}$$ or $$\ce{IBr}$$ with double bonds is slow. You have to wait half an hour for the reaction to be finished. But of course, you may wait a longer time, maybe $$\pu{3 h}$$ as you suggest. Or one night. Usually we are in a hurry, and we are looking for ways of avoiding to spend too much time for a given operation. That is why the procedure recommends half an hour.
And of course you may carry out the titration with a solution 5 times more diluted than recommended. The only drawback is that you may have to refill many times your burette before going to the end of the titration. Suppose $$\pu{42 mL}$$ of the $$\pu{0.1 M}$$ solution is required for performing the titration. This could be done with a usual $$50$$-$$\pu{mL}$$ burette. With a $$\pu{0.02 M}$$ solution, you will use $$5 \times \pu{42 mL} = \pu{210 mL}$$. In this case you will have to refill four times your burette. Four times! Bad luck!