Question

A solution was prepared by mixing 10.00 mL of unknown (X) with 5.00 mL of standard (S) containing 8.24 μg S/mL and diluting to 50.0 mL. The measured signal quotient (signal due to X/signal due to S) was 1.69. In a separate experiment, it was found that, for equal concentrations of X and S, the signal due to X was 0.930 times as intense as the signal due to S. Find the concentration of X in the unknown. Please show work.

Answer 1

It is given that signal X is 0.93 X signal S[ with same concentrations]

Let us assume that the concentration of X = x

the concentration of [S] = s

also, signal due to X / signal due to S = 1.69

So, 0.93 x / s = 1.69 [Beer lambert's law]

x / s = 1.82

Mass of "S" = Mass present in 1mL X total volume taken = 8.24 ug/ml x 5 ml = 41.2 ug

concentration of S in 50mL = Mass / total volume = 41.2 / 50 = 0.824ug /ml

therefore x = 1.82 X s = 1.82 X 0.824 = 1.499ug /ml

x = Mass of X / total volume = 1.499 ug /ml

Mass of X = 1.499 X 50 = 74.95ug

This mass is obtained by initial additon of 10mL of X

Hence the concetnration of [X] = 74.95 ug / 10mL = 7.495ug / mL