mk_2007 Posted November 5, 2008 Share Posted November 5, 2008 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) Link to comment Share on other sites More sharing options...
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