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standard preparation
Posted: Wed Aug 03, 2016 5:24 pm
by sriharshabio
Hello,
I have a stock solution concentration of 1 mg/ml for 18 individual standards. i need to prepare a mixture of standards of these 18 stds where the compounds need to be mixed at a concentration of 10 microgram/ml and i need a final volume of 10 ml of this mix solution. Could somebody please elaborate the calculation. Thanks.
Re: standard preparation
Posted: Wed Aug 03, 2016 5:48 pm
by Don Shelly
10 microgram per ml equals .01 mg/ml.
Multiply .01 times FV of 10 ml = 0.1 mg
0.1 mg times 1 mg/ ml = 0.1 ml or 100 microliters
Re: standard preparation
Posted: Wed Aug 03, 2016 5:58 pm
by sriharshabio
Hi Don,
Thanks for the reply. So if i add 100 microlitres of 1 mg/ml of all the 18 stds into a tube i will get final mixture concentration of 10 microgram/ml. Is that right?
Regards,
Sri
Re: standard preparation
Posted: Wed Aug 03, 2016 6:04 pm
by Don Shelly
Hi Don,
Thanks for the reply. So if i add 100 microlitres of 1 mg/ml of all the 18 stds into a tube i will get final mixture concentration of 10 microgram/ml. Is that right?
Regards,
Sri
and take to a 10 ml final volume with solvent.
Re: standard preparation
Posted: Wed Aug 03, 2016 6:10 pm
by sriharshabio
ok thank You
Regards,
Sri
Re: standard preparation
Posted: Wed Aug 03, 2016 6:14 pm
by GOM
Hi
Another way of expressing it is
1mg/ml = 1x10-3g/ml = 1x10-4g/100ul
100ul diluted to 10ml
=1x10-4g in 10ml
=1x10-5g in 1ml
0r 10x10-6 in 1ml
=10ug/ml
So yes, 100uL of each - a total 1.8mls made up 10mls will give 10ug/ml
But you really should be capable of working this out for yourself
Regards
Ralph