1. Show your calculations for preparing the following solutions: 200 ml of 20% (
ID: 980969 • Letter: 1
Question
1. Show your calculations for preparing the following solutions: 200 ml of 20% (w/v) NaOH, 1 liter of 1.0 M Tris [molecular weight (MW) 121.1 g/mol], and 100 ml of 0.2 M EDTA (MW 372.2 g/mol).
How much of the above Tris (1M) and EDTA (0.2M) solutions are used to prepare a 100 ml of TE buffer (10mM Tris and 1 mM EDTA)?
Show your calculations for preparing the following solutions: 200 ml of 20% (w/v) NaOH, 1 liter of 1.0 M Tris [molecular weight (MW) 121.1 g/mol], and 100 ml of 0.2 MEDTA (MW 372.2 g/mol. How much of the above Tris (1M) and EDTA (0.2M solutions are used to prepare a 100 ml of TE buffer (10mM Tris and 1 mMEDTA)?Explanation / Answer
(a) 200ml of 20% NaOH
Mass of NaOH needed = (20/100) x 200 = 40g
40g of NaOH is needed to be dissolved in 200 ml of solvent to get 20% NaOH solution
(b) 1 L of 1.0M Tris
Molarity = No of moles of solute/ Volume of solution (in L)
No. of moles = Mass of given substance(m) / Molecular mass of the substance(M)
1.0 = m/121.1/1
m = 121.1g
Therefore 121.1 g of Tris is needed to prepare 1L of 1.0M solution.
(c) 100ml of 0.2 MEDTA
Molarity = No of moles of solute/ Volume of solution (in L)
No. of moles = Mass of given substance(m) / Molecular mass of the substance(M)
0.2 = m/372.2/0.1
m = 7.44g
Therefore 7.44g of EDTA is needed to prepare 100ml of 0.2M EDTA solution.
Tris = 10mM = 0.01M
EDTA = 1mM = 0.001M
For 1L TE buffer, 100ml of Tris is required and 20 ml of 0.2 M of EDTA is required
Therefore for 100ml of TE buffer 10ml of tris and 0.4 ml of EDTA is required.
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