Answer to Question #109275 in General Chemistry for rodaina

Question

17. A tenfold serial dilution is performed with a final dilution of 1/10000. The beginning dilution is the series in tube 1 is a 1/10 dilution. If the original concentration had been 25000 ng/dL, what would the concentration be in tube 3?

Answer 1

1 dL =0.1 litres

original concentration=25000ng/dL = 25000$\times$10^-9/0.1 g/L =2.5$\times$10^-4 g/L

in the first tube it is diluted to 1/10 of the original hence the concentration will become

$\frac{1}{10}\times2.5\times10^{-4}=2.5\times10^{-5}g/L$

similarly in the 2nd second tube it would 1/10 of the previous concentration which will be

$\frac{1}{10}\times2.5\times10^{-5}=2.5\times10^{-6}g/L$

similarly in the 3rd second tube it would 1/10 of the previous concentration which will be

$\frac{1}{10}\times2.5\times10^{-6}=2.5\times10^{-7}g/L$