Solution I consists of 50 mM glucose, 10 mM EDTA, and 25 mM Tris base. Starting
ID: 217223 • Letter: S
Question
Solution I consists of 50 mM glucose, 10 mM EDTA, and 25 mM Tris base. Starting with glucose (a solid, MW = 180 g/mole), EDTA (0.5 M solution), and Tris base (a solid, MW = 121 g/mole), calculate the quantities you would need to make 100 ml of Solution I.
Solution II consists of 0.2 N NaOH and 1% SDS (sodium dodecyl sulfate). Starting with sodium hydroxide (a solid, MW = 40 g/mole) and SDS (a solid, MW = 288 g/mole), calculate the quantities you would need to make 50 ml of Solution II.
Solution III is 7.5 M ammonium acetate. Ammonium acetate is a solid and its MW is 77 g/mole. How much ammonium acetate would you need to make 150 ml of Solution III Solution III is 7.5 M ammonium acetate. Ammonium acetate is a solid and its MW is 77 g/mole. How much ammonium acetate would you need to make 150 ml of Solution III
The recipe for one liter of 5X TBE is 0.45 M Tris base, 0.45 M boric acid, and 0.01 M EDTA. Referring back to your answer in part “a”, how would you make the appropriate amount of 10X TBE? Note: Tris base is a solid and has a MW of 121 g/mole, and that you have a 0.5 M EDTA stock solution. Boric acid is a solid and has a MW of 62 g/mole.
Explanation / Answer
Solution I consists of 50 mM glucose, 10 mM EDTA, and 25 mM Tris base.
Starting with glucose (a solid, MW = 180 g/mole), EDTA (0.5 M solution), and Tris base (a solid, MW = 121 g/mole), calculate the quantities you would need to make 100 ml of Solution I.
Answer:
50 mM glucose preparation:
Grams=Molarity x (Volume x Molecular weight)
=0.05 M x 0.1 L x 180
=0.9g of glucose
25 mM Tris base
=0.025 M x 0.1 L x 121
=0.3g of tris base
10 mM EDTA
Stock concentration= 0.5M
Required concentration= 0.01M
Required Volume= 0.1L
C1V1=C2V2
(Stock) (required)
V1= C2V2/C1
V1= (0.01M x 0.1L)/0.5M = 2mL
Take 2 mL of 0.5M EDTA + 0.9g of glucose+0.3g of tris base and make up the volume to fimal 100 mL. Solution I is ready
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Solution II consists of 0.2 N NaOH and 1% SDS (sodium dodecyl sulfate). Starting with sodium hydroxide (a solid, MW = 40 g/mole) and SDS (a solid, MW = 288 g/mole), calculate the quantities you would need to make 50 ml of Solution II.
Answer:
0.2 N NaOH preparation:
Grams=Molarity x (Volume x Molecular weight)
=0.2 M x 0.05 L x 40 g/mol
=0.4g of NAOH
1% SDS
0.5g
The above two weighed and dissolved in a total of 50 mL volume gives solution II
Solution III is 7.5 M ammonium acetate. Ammonium acetate is a solid and its MW is 77 g/mole. How much ammonium acetate would you need to make 150 ml of Solution III
Answer:
7.5 M ammonium acetate
Grams=Molarity x (Volume x Molecular weight)
=7.5 M x 0.15 L x 77g/mol
=86.6g of ammonium acetate dissolved in total of 150mL water gives solution III
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