a) Stray light in the spectrophotometer = 0.421 - 0.827/2 =
0.0075
b) For the given data
x g of compound in 5 ml gave a solution
1 ml of above solution diluted to 10 ml gave an absorbance =
0.827
concentration of compound = absorbance/molar absorptivity
                                          Â
= 0.827/2.130 x 10^4
                                          Â
= 3.88 x 10^-5 M
concentration in original 5 ml sample solution = 3.88 x 10^-5 x
10
                                                                       Â
= 3.88 x 10^-4 M
c) Mass of compound used to make 5 ml solution = 3.88 x 10^-4 M
x 0.005 L x 292.16 g/mol
                                                                              Â
= 5.67 x 10^-4 g
                                                                              Â
= 0.57 mg