Map Sapling Learning Consider a solution containing 4.63 mM of an analyte, X, an
ID: 1059683 • Letter: M
Question
Map Sapling Learning Consider a solution containing 4.63 mM of an analyte, X, and 1.1 9 mM of a standard, S. Upon chromatographic separation of the solution peak areas for X and S are 3913 and 10695, respectively. Determine the response factor for X relative to S. Number To determine the concentration of X in an unknown solution, 1.00 mL of 8.25 mM S was added to 5.00 mL of the unknown X solution and the mixture was diluted to 10.0 mL. After chromatographic separation, this solution gave peak areas of 6065 and 4849 for Xand S, respectively. Determine the concentration of S in the 10.0 mL solution. Number mM Determine the concentration of X in the 10.0 mL solution. Number mM Determine the concentration of X in the unknown solution. O Previous ® Give Up & View Solution Check Answer Next Ext HintExplanation / Answer
From the data
Response factor = (area/conc)x/(area/conc(s)
= (3913/4.63)/(10695/1.19)
= 845.14/8987.395
= 0.094
Determination of concentration of unknown X
concentration of S in 10 ml solution = 8.25 mM x 1 ml/10 ml = 0.825 mM
So,
concentration of X in 10 ml solution = 6065 x 0.825/0.094 x 4849 = 10.98 mM
concentration of X in unknown 5 ml solution = 10.98 x 10/5 = 21.96 mM