Answer to Question #87004 in General Chemistry for Lim Yoona

Question

For the following runs the concentration of the dyes were calculated as per below:

RUN [DYE] (mol L-1)
1 1.09 X 10-6
2. 2.43 X 10-6

Based on the above measurements the ln[DYE] for run 2 will be:

Answer 1

ln[1.09 x 10-6]= 13.73

ln[2.43 x 10-6] = 12.93