(c) Calculating concentration
The formula for absorbance #A# is given by Beer's Law:
#color(blue)(bar(ul(|color(white)(a/a)A = epsiloncl color(white)(a/a)|)))" "#
where
#epsilon =# the molar absorptivity of the sample
#c =# the concentration of the sample
#l =# the path length through the cuvette
It states that the absorbance is directly proportional to the concentration of the sample. Thus,
#A_2/A_1 = (color(red)(cancel(color(black)(epsilon)))c_2color(red)(cancel(color(black)(l))))/(color(red)(cancel(color(black)(epsilon)))c_1color(red)(cancel(color(black)(l)))) = c_2/c_1#
We can rearrange this formula to give
#c_2 = c_1 × A_2/A_1#
In your problem,
#c_1 = 1.0 ×10^"-3"color(white)(l) "mol/L"; A_1 = 0.782#
#c_2 = ?; color(white)(mmmmmmml)A_2 = 0.553#
∴ #c_2 = 1.0 × 10^"-3"color(white)(l) "mol/L" × 0.553/0.782 = 7.1 × 10^"-4"color(white)(l)"mol/L"#
(d) Determining the new setup
You know that
#A = epsiloncl#
#epsilon# is a constant that is characteristic of the sample.
You want to keep #A_0# constant at 0.782.
Thus, if you double the path length #l#, you must halve the concentration #c#.
Start with
#c_0 = 1/2 × 1.0 × 10^"-3"color(white)(l) "mol/L" = 5.0 × 10^"-4"color(white)(l)"mol/L"#