Calculations

Question 1

Determine the volumes/amounts of each of the reagents you would combine in order to generate the following:
0.3 M solution of NaCl. Atomic weights: Na, 23.0 g.mol-1; Cl, 35.5 g.mol-1. Assume you require 750 mL final volume

Answer 1

mass = conc (m/L) x vol (L) x mol weight (g/mol)
= 0.3 x 0.75 x (23 + 35.5)
=3.16g

Question 2

Determine the volumes/amounts of each of the reagents you would combine in order to generate the following:
0.3 M solution of NaCl. Atomic weights: Na, 23.0 g.mol-1; Cl, 35.5 g.mol-1. Assume you require 125 mL final volume

Answer 2

mass = conc (m/L) x vol (L) x mol weight (g/mol)
= 0.3 x 0.125 x (23 + 35.5)
=2.19g

Question 3

Determine the volumes/amounts of each of the reagents you would combine in order to generate the following: 35 mM solution of Tris base, assuming you require 1.5 L final volume. Molecular weight (MW) of Tris is 121.1 g.mol-1.

Answer 3

mass = conc (m/L) x vol (L) x 121.1

=6.35g

Question 4

Determine the volumes/amounts of each of the reagents you would combine in order to generate the following: 15% v/v solution of glycerol (MW 92.09 g.mol-1), assuming you require 0.25Lfinal volume.

Answer 4

15% x 0.25 = 0.0375L glycerol = 37.5mL

0.25 - 0.0375 = 0.2125L water = 212.5mL

Question 5

Determine the volumes/amounts of each of the reagents you would combine in order to generate the following: 25% w/v solution of sucrose (MW 342.30 g/mol), assuming you require 50 mL.

Answer 5

25% x 50mL = 12.5mL = 12.5g sucrose

50mL - 12.5mL = 37.5 mL water

Question 6

Determine the volumes/amounts of each of the reagents you would combine in order to generate the following:
750 mL of a 20% NaCl solution (MW NaCl: 58.5g/mol )

Answer 6

20% x 750mL = 150mL NaCl = 150g NacL

Make up to 750mL with water

Question 7

Determine the volumes/amounts of each of the reagents you would combine in order to generate the following:
250 mL of a 10X TE stock (Tris, EDTA). 1X TE is 10 mM Tris and 1 mM EDTA. The MW of EDTA is 292.24 g.mol-1.

Answer 7

EDTA
mass = conc (m/L) x vol (L) x mol weight (g/mol)
= 0.01M x 0.25L x 292.24
=0.731g

Tris
mass = conc (m/L) x vol (L) x mol weight (g/mol)
= 0.1M x 0.25L x 292.24
=7.316g

Add both and make up to 250mL with water

Question 8

Determine the volumes/amounts of each of the reagents you would combine in order to generate the following:
600 mL of the following stock solution: 0.6 M Tris, 60 mM SDS (sodium dodecyl sulphate is a denaturing detergent; MW 257 g.mol-1), 1% (w/v) powdered milk (MW 49 g.mol-1), 1.3 M NaCl (MW 58.8 g.mol-1).

Answer 8

Tris
mass = conc (m/L) x vol (L) x mol weight (g/mol)
= 0.6M x 0.6L x 121.1g/mol
=43.596g

SDS
mass = conc (m/L) x vol (L) x mol weight (g/mol)
= 0.06M x 0.6L x 257g/mol
=9.25g

Powdered milk
w/v given as 1%
mass =percentage x total vol of solution
          =0.01 x 600mL
          =6g
NaCl
mass = conc (m/L) x vol (L) x mol weight (g/mol)
= 1.3M x 0.6L x 58.8g/mol
=45.86g

Add all and make up to 600mL with water

Question 9

How much SDS (sodium dodecyl sulfate, MW=288,372 g.mol-1) do you need to make up 200 mL of a 20% solution?

Answer 9

20% x 200mL =40mL

Question 10

Determine the quantities of each of the reagents would you require in order to generate the following:

25% v/v solution of bleach, assuming you require 250 ml total volume. (Sodium hypochlorite has a MW of 74.44 g/mol)

Answer 10

25% x 250mL = 62.5mL sodium hypochlorite

Make up to 250ml with water

Question 11

Determine the quantities of each of the reagents would you require in order to generate the following:
2.2 M solution of NaCl. Atomic weights: Na, 22.9 g.mol-1; Cl, 35.5 g/mol. Assume you require 100 mL final volume

Answer 11

mass = conc (m/L) x vol (L) x mol weight (g/mol)
=2.2M x 0.1L x (22.9 + 35.5)
=12.848g

Question 12

Determine the quantities of each of the reagents would you require in order to generate the following:
25% w/v solution of fructose (MW 180.2 g/mol), assuming you require 0.25L

Answer 12

25% x 0.25L = 0.0625L = 62.5mL = 62.5g fructose

Make up to 0.25L with water

Question 13

How would you make a 120 mL agarose gel of 0.6% (w/v) in 0.5X TBE buffer. The TBE (Tris, Borate, EDTA) stock is 5X and MW agarose is 210 g.mol-1

Answer 13

TBE stock
We want 0.5X. Stock is 5X. Hence dilution factor is 1/10
1/10 x 120mL = 12mL TBE stock

Agarose
0.6% x 120mL = 0.72mL

Question 14

You are preparing a 40 mL agarose gel of 0.8% (w/v) in 0.5X TBE buffer. The TBE (Tris, Borate, EDTA) stock is 10X; the molecular weight of agarose is 210 g.mol-1. Determine the amounts of each reagent required:

Answer 14

TBE stock
We want 0.5X. Stock is 10X. Hence dilution factor is 1/20
1/20 x 40mL = 2mL TBE stock

Agarose
0.8% x 40mL = 0.32g

Water
40mL - 2mL - 0.32mL = 37.68mL

Question 15

You have a 100X stock of buffer and you need 150 µL of 1X. How do you make this?

Answer 15

We want 1X buffer. Stock is 100X buffer. Hence dilution factor is 1/100
1/100 x 150uL = 1.5uL stock
Make up to 150uL with diluent

Question 16

How would you prepare 3X TBE buffer from a 12X TBE buffer for a total volume of 200 mL?

Answer 16

Dilution factor: 3X/12X = 1/4
1/4 x 200mL = 50mL
Make up to 200mL with water

Question 17

You need to prepare a 250 mL solution comprising 25 mM Tris, 25 mM EDTA, and 0.1 M NaCl. What volumes (mL) would be required from the stocks of 1 M Tris, 0.5 M EDTA, 750 mM NaCl, and dH2O?

Answer 17

Tris
Dilution factor: want 0.025 M. Have 1M. Hence 0.025
0.025 x 250mL = 6.25mL

EDTA
Dilution factor: want 0.025M. Have 0.5M. Hence 0.025/0.5=0.05
0.05 x 250mL = 12.5mL

NaCl
Dilution factor: want 0.1M. Have 0.75M. Hence 0.1/0.75=0.1333
0.1333 x 250mL = 33.33mL

dH2O
250 - (6.25 + 12.5 + 33.33) = 197.92mL