Question

4. The calibration data for the determination of lead levels in fruit juices by a differential pulse polarography method is shown below. 40 10 30 20 50 Current /HA0.0000.1180.2400.5070.7130.9811.183 When these data were plotted using linear regression a correlation coefficient, R2, of 0.9988 and a line of best fit were obtained; y-0.0239 x + 0.005 Use this information to determine the lead levels in the unknown sample, which had been diluted by a factor of 25, from the following replicate measurements; 0.475, 0.481, 0.486, 0.479 and 0.484.

Hi can someone please show how this is solved, I have the answer as = 497.9 +/- 0.73 ngL-1 but would just like to know how the answer was reached. Especially to the +/- 0.73 ngL-1

Thank you.

Answer 1

Solution:

Given that

Sol 507 30 0·チ13 50 1-183

abie ead levellp anbe turd as: x 66s X 19.916

ス= &0.041 al le , Xay 19.9153 m NDTE annhali dilit diet n)