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)