How would you (literally) make a 1.5L stock of 10x Western Transfer Buffer, which contains 0.11...
How would you (literally) make a 1.5L stock of 10x Western Transfer Buffer, which contains 0.11 M tris-base and 2.35 M glycine? The molecular weight of tris-base is 121.14 g/mol and the molecular weight of glycine is 75.07 g/mol.
Homework Answers
the volume of the buffer needed= 1.5 L
molarity of Tris-Base= 0.11 M
molarity= number of moles/volume of the solution in liters
number of moles= mass in gram/gram molecular mass
Molarity= 0.11 M
0.11 M= number of moles/ 1.5 L
number of moles= 0.11 
1.5
= 0.165
number of moles= mass in gram/ gram molecular mass
0.165= mass in gram/121.14
mass in gram= 0.165121.14
= 19.9881 g
Glycine
Molarity= 2.35 M
Molarity=number of moles/ volume of the solution in liters
2.35= number of moles/ 1.5
number of moles= 2.35
1.5
=3.525
number of moles=mass in gram/gram molecular mass
3.525 = mass in gram/75.07
mass in gram= 3.52575.07
= 264.62175 g
take 264.62175 g of glycine and 19.9881 g of Tris base and add water and dissolve the solutes and add water such that final volume=1.5 L
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How would you (literally) make a 1.5L stock of 10x Western
Transfer Buffer, which contains 0.11...
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