Changing Concentration

Question

If 25 ml of creatinine (c4h7n3o) stock standard (conc= 1.0 g/dl) are diluted to 100ml to make an intermediate standard and then 15ml of the intermediate are diluted to 300ml to make a working standard, what is the concentration of the latter in mg/dl?

Answer 1

There are several ways this can be done.  Here is the simplest.

 

Calculate the OVERALL dilution factor:

25 ml to 100 ml = 4 fold dilution (100/25)

15 ml to 300 ml = 20 fold dilution (300/15)

Overall dilution factor = 4 x 20 = 80 fold dilution

Final concentration starting with 1.0 g/dl and applying an 80 fold dilution factor = 1/80 = 0.0125 g/dl

0.0125 g/dl x 1000 mg/g = 12.5 mg/dl

 

Doing it by a more extended procedure (and there are many ways to do this as well):

1g/dl = 1 g/100 ml = 1000mg/100 ml = 10 mg/ml STOCK solution

First dilution:  10mg/ml x 25 ml = 250 mg put into 100 mls = 250 mg/100 ml

Second dilution:  250 mg/100 ml x 15 ml = 37.5 mg put into 300 ml = 37.5 mg/300 ml FINAL SOLUTION

Convert this to desired units of mg/dl:  37.5 mg/300 ml = 0.125 mg/ml;  0.125 mg/ml x 100 ml/dl = 12.5 mg/dl