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Calculating solutions.. volumes required..


mk_2007

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Hi! basically, I'm doing some practice questions, and came across one, which i'm not quite sure how to calculate.. I had a go at it and this is my answer.. can anybody tell me if the method I used is correct, or if I'm going about this the wrong way!!

 

The question was.. a lysis solution contained

50 mM Tris, 100mM NaCL and 0.01% (w/v) SDS. How would you prepare 100 ml of this solution?

 

So what I did was :

 

Start Final

Volume x Concentration= Volume x concentration

 

 

So Taking a start concentration of 1 M, then

TRIS = T x 1M = 100 ml x 50 x 10-³ M

NaCl = N x 1M = 100 ml x 100 x 10-³ M

SDS= 0.01% is the same as 0.01 g/100ml. Therefore this can be taken as the concentration.

 

So Volumes of each concentration of solutions required to make up 100 ml are TRIS= 5ml

NaCl= 10 ml

SDS= 0.01 ml

H2O = 84.99 ml (100-5+10+0.01)

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