1
$\begingroup$

A student wants to make serial dilution of a sample. She starts with a 1mg/ml sample and 10 tubes, (labelled 1-10) each containing 2 ml buffer.

The student puts 1ml of sample solution into tube 1 Repeating the process.

What is the concentration of the sample in tube 4?

Give your answer to 2 decimal places and in µg/ml

I've started this question and I assume I would do 1/2^4 which would give me 0.0625

But i'm stuck if this is right?

$\endgroup$
3
$\begingroup$

I think 1/3^4 is the right factor which gives 12.3 microgm/ml step #1 consists in adding 1ml of the sample to 2 ml of buffer in tube #1 which then contains 1mg in 1+2=3 ml's i.e has concentration 1/3 mg/ml.Repeating we get 1/3^2 in tube #2 and ...1/3^k in the kth tube. for k=4 we get 1/81 mg/ml=1000/81 microgram/ml As @xusr remarked this line of reasoning is a good first order approximation based on the additivity of volumes.If one needs a more exact answer in general e.g for mixtures of organic and aqueous buffers etc one must consider the partial molar volumes of the components involved which will be be dependent both on their chemical nature and the thermodynamic state parameters of the experiment leading to nonadditivity of the volumes mixed or in worst case to phase separations(i.e solute is present only in one portion of the resultant solution)

$\endgroup$
  • 3
    $\begingroup$ I think you are correct, could you please show your work for the serial dilutions. I think it would be beneficial for @user135629 to have your work shown $\endgroup$ – JSCoder says Reinstate Monica Oct 31 '17 at 19:19
  • $\begingroup$ OK @xusr.Of cource if one wants to be very exact one has to consider the $\endgroup$ – Dikeos Mario S. Nov 2 '17 at 11:11
  • $\begingroup$ @xusr.I have discussed the matter completely in principle by adding to my answer what has to be considered in general when the first approximation I $\endgroup$ – Dikeos Mario S. Nov 4 '17 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.