Hi good afternoon to all of you! i would like to resolve my clarification with you that i had 93.7 mg of protein concentration present in 1 ml of sample and i need 1 mg for 50 ml so how do i prepared this one...
take 0,533 ml of stock solution and add to 49,467 ml of necessary buffer
Stock concentration = 93.7mg/mL
i.e. 93.7mg in 1000 microlitre
i.e. 9.37mg in 100 microlitre
i.e. 0.937mg in 10 microlitre
i.e. 1 mg in 10.6 microlitre
Hence you can take 10.6 microlitre from stock and add into 49 mL and 989.4 microlitre of buffer/water.
v1c1=v2c2
v1=v2c2/c1=(50 ml*1 mg/ml)/ 93.7 mg/ml = 0.5336 ml
Required vol= 50 ml
So, 0.5336 ml of stock dilute with 49.4664 ml of solvent or buffer
thank you so much sir !
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