Question

The following chemicals have the indicated formula weights:

Sucrose -- 342.3g/mol

Potassium dihydrogen phosphate (KH2P04) -- 136.09g/mol

You also have the following solutions already made: 20% SDS, 5M NaCl and 10X TAE List the steps to make the following buffers (be specific and show calculations):

a. 1 liter of a solution containing: 0.5 M sucrose, 0.05 M NaCl, 0.02 M KH2PO4.

b. 100 ml of 10% SDS, 1M NaCl.

c. 40 ml 0.5% agarose in 1X TAE solution (calculate for both agarose and 1X TAE).

Answer 1

a)

Sucrose formula weight-342.3g/mol, as formula weight means total sum of the atomic weight of a particular compound,

thus 1 mole of sucrose has the mass corresponding to 342.3gm

to make 1 (M) , 1 lt solution we need 342.3 gm sucrose

thus for 0.5(M), 1 lt solution, we need 342.3 X 0.5 /1=171.15gm of sucrose

We already have 5M Nacl solution, so to make 0.05M Nacl solution we need to dilute the stock solution 5/ 0.05=100 fold.

for 1 lt solution thus we need to add 1lt/ 100 or 1000ml/ 100 =10 ml of 5M Nacl solution.

formula weight of KH2P04 is   136.09g/mol

Thus for 1M , 1lt solution we need to dissolve 136.09gm KH2P04

for 0.02M , 1lt solution we need 136.09 X 0.02/1= 2.7218 gm KH2P04

to prepare 1 liter of a solution containing: 0.5 M sucrose, 0.05 M NaCl, 0.02 M KH2PO4., we need 171.15gm of sucrose+10 ml of 5M Nacl solution + 2.7218 gm KH2P04 + 990ml of water.

b) SDS stock solution is 20%,

so for 100ml 10% solution we need to dilute the SDS 20/ 10= 2 fold

so we required 100/2= 50 ml of SDS from the stock solution.

NaCl stock solution is 5M, so to make 1M NaCl, we need to dilute it 5/1= 5 fold

thus for 100ml of solution we require 100/5= 20 ml of Nacl from 5M NaCl

Thus to prepare 100 ml of 10% SDS, 1M NaCl -50ml 20% SDS+20ml Nacl (5M)+30ml water.

c) TAE stock solution is 10X, so to make 1X TAE buffer we require to dilute 10/1=10 fold,

thus to make 40ml TAE 1X buffer, we need to take 40/ 10= 4 ml from 10X TAE stock and add 46 ml of water to it.

Now for 0.5% agarose,

100ml solution require 0.5gm agarose

40ml solution require 0.5 X 40/ 100=0.2 gm agarose

so, to prepare 40 ml 0.5% agarose in 1X TAE solution,

4 ml from 10X TAE stock + 46 ml of water + 0.2 gm agarose