If it takes #"[HALF of the]"# Iodine-131 approximately 8 days to change to Xenon-131, how many days would have passed if 75% of it has changed to Xenon-131?

Well, it took #16# days, two half-lives, if #t_"1/2" = "8 d"#... actually, with the true half-life, it would be #16# days and #56.7# minutes.

The nuclear reaction is a beta decay:

#""_(53)^(131) "I" -> ""_(54)^(131) "Xe" + ""_(-1)^(0) e#

This is first order with respect to #""_(53)^(131) "I"#, which follows:

#[""_(53)^(131) "I"] = (1/2)^(t//t_"1/2") [""_(53)^(131) "I"]_0#

where #n = t//t_"1/2"# is the number of half-lives.

If it takes #8# days for a FULL conversion as the question says, an INFINITE number of half-lives occurred such that:

#0 = (1/2)^(n) [""_(53)^(131) "I"]_0#

And this would give an infinite number of solutions to how many days it "should" take to go through two half-lives.

So this question has a major typo... It SHOULD be that it takes #8# days for HALF of the #""_(53)^(131) "I"# to turn to #""_(54)^(131) "Xe"#...

Hence, we really mean that:

#ul(t_"1/2" = "8 d")#

This means that losing #75# of the #""_(53)^(131) "I"# would lead to:

#[""_(53)^(131) "I"] = (1 - 0.75)[""_(53)^(131) "I"]_0 = (1/2)^(t//8)[""_(53)^(131) "I"]_0#

Thus,

#0.25 = (1/2)^(t//8)#

#ln 0.25 = t/8 ln (1/2)#

#color(blue)(t) = (8ln 0.25)/ln (1/2) = color(blue)ul("16 d")#

or 16 days.

That makes sense because clearly, #1/4 = (1/2)^2#, so #n = 2# half-lives passed by to reduce the concentration to #1/4# (or by #3/4#)...