20% SDS 10ml_grams of SDS is needed 0 5 M NaOH 30 ml_grams of NaOH is needed 1 M

Question

20% SDS 10ml_grams of SDS is needed 0 5 M NaOH 30 ml_grams of NaOH is needed 1 M Tris Cl, pH 7.6 50 ml(0.45mu m filter) (One can obtain the molecular weight of Tris online)_grams of Tris is needed 1.5 M Tris Cl pH8 8 50 ml(0 45pm filter)_grams of Tris is needed 0.5 M Tris Cl pH6.8 50 ml(0.45pm filter)_grams of Tris is needed 20% Ethanol 200 ml_ml of 100% ethanol will be added into_dH2O 4 M NaCl 50 ml(0 45mu m filter)_grams of NaCl is needed Imidazole 3M 10 ml(0.45mu m filter)_grams of imidazole is needed Then use the stock solutions made above to make the following solutions Buffer 1 200 ml 20mM Tris. Cl pH7.6 5mM imidazole. 100mM NaCl_ml of 1 M Tris. Cl pH7.6, _ml 3 M imidazole, _ml 4 M NaCl ml dH_2O Buffer 2 100 ml 20mM Tris Cl pH7.6. 10mM imidazole. 100mM NaCl_ml of 1 M Tris. Cl pH7.6, _ml 3 M imidazole, _ml 4 M NaCl, _ml dH_2O Buffer 3 100 ml 20mM Tris. Cl pH7.6. 20mM imidazole. 100mM NaCl_ml of 1 M Tris Cl pH7.6._ml 3 M imidazole, _ml 4 M NaCl_ml dH_2O Buffer 4 100 ml 20mM Tris Cl pH7.6 200mM imidazole. 100mM NaCl_ml of 1 M Tris Cl pH7.6, _ml 3 M imidazole._ml 4 M NaCl._ml dH_2O To make buffer 1-4. please use M_1V_1 = M_2 V_2. to figure out how much of stock solutions one needs For the rest of the volume, use dH_2O.

Explanation / Answer

Q3.

TRIS compound = 121.14 g/mol

then, if we require 1 M of Tris compound

M = mol/V

mol = mass/MW

so

M = mass/(MW*V)

substitute data

mass = M*MW*V = 1*121.14*50/1000 = 6.057 g of TRIS compound required

Q4.

Similar, we nened

mass = M*MW*V

mass of TRIS = 1.5*121.14*50/1000 = 9.0855 g

Q5.

Same case as in Q3 and 4

Similar, we nened

mass = M*MW*V

mass of TRIS = 0.5*121.14*50/1000 = 3.0285 g

Q6.

20% ethanol, assume this basis is per volume

so

v/v% = volume of ethanol / total volume *100%

we need 20% so

20%/100% = x /200

x = 0.2*200 = 40 mL of Ethanol reqruied

Q6.

4M of NaCl in V = 50 mL

note that

MW of NaCl = 58.440 g/mol

then

M = mol/V

M = mass/(MW*V)

mass = M*MW*V

mass of NaCl = 4*58.44*50/1000 = 11.688 g of NaCl needed

Q7.

M = 3 M

MW of imidazole = 68.077 g/mol

V = 10 mL

now, same as in previous steps

M = mass/(MW*V)

mass = M*MW*V

mass = 3*68.077*10/1000

mas of Imidazole = 2.04231 g

For further multiple questions, pleas post them in different set of Q&A.

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