Absorbance is
not 1/transmittance. Absorbance is -log
10(transmittance); transmittance = 10^-(absorbance).
either my method for solving the first question was wrong and I just got the correct answer by chance
Yup. Pure coincidence.
I simply took 0.100 and multiplied it by 0.9 (which I got from 90% which I got just by doing 100%-10%).
Why? What's the physical basis of it? Do you understand what this means, in terms of what's physically going on in the spectrometer? (I had to think about it for a while, because I don't regularly do AA.)
Suppose that a monochromator also passes a neon fill gas line at 588.2nm that has an intensity of 10% of the sodium line when no absorbance occurs
Here's what I think it means. The spectrometer is passing light from a sodium source, plus the neon line, through the empty sample chamber. With no absorption, the total light intensity reaching the detector is
I
0(apparent) = I
0(Na) + I(Ne) = I
0(Na) + 0.1I
0(Na) = 1.1I
0(Na)
With a sample containing enough Na to give a (true) absorption of 0.1, the fraction of Na light transmitted is 10
-0.1 = 0.794. So
I(apparent) = 0.794I
0(Na) + 0.1I
0(Na) = 0.894I
0(Na)
So the apparent transmission is I(apparent)/I
0(apparent) = 0.894/1.1 = 0.813
and the apparent absorbance is -log(0.813) = 0.090.
Now try it with true absorbance = 1.0 (It works).