Here is the problem along with the answer. Can you please explain to me how they found Xi?

1. Triple-quad GC/MS was recently used to analyze a sample of water from the Ohio River for 2,2',4,4'- Tetrabromodiphenyl ether (2,2',4,4' TBDE), a toxic component with very, very low water solubility. Analysis gave a signal area of 103.5 for one sample 2,2,4,4 TBDE. A spike of 5.00L of 3.00mM2,2,4,4 TBDE was added to 95.00L this solution. The final spiked solution gave a 2,2',4,4' TBDE signal area of 259.0. Determine the concentration of 2,2,4,4 TBDE in the original water sample in ppmw/v. (2,2,4,4 TBDE formula mass =485.79g/mol) (SHOW ALL CALCULATIONS and UNITS). Xf=[Xi](VVo)[Sf]=[Si](VVs)[Sf]+[Xf][Xi]=IS+XIXXf=[Xi](100.00mL95.00mL) [Sf]=[3.00103M](100.00L5.00L)=1.50104M [1.50104M]+[Xi](100.00mL95.00mL)[Xi]=259.0nA103.5nA=3.996101 [Xi]=9.66105M2,2,4,4TBDE