Question

In this activity, you will use solutions math to prepare recipes to make a number of solutions. A list of available substances to use when preparing your solutions is included below, only those solutes or solutions can be utilized.

In the cabinet you have the following substances:

A bottle of stock 90% Ethanol

A bottle of stock 1M NaHCO₃ (liquid; mw=84.07)

A bottle of sucrose crystals (solid; mw=342)

A large bottle of NaCl (solid; mw=58)

A large bottle of agarose (solid; mw=306)

1 bottle of NaH2PO4 (solid; mw=120)

1 bottle of Na2HPO4 (solid; mw=142)

Bottle of estradiol (solid; mw= 272.38)

A bottle of stock 3 M PBS buffer (liquid;

A bottle of stock TBS buffer (liquid)

Huge jug of TAE (liquid)

Huge jug of distilled water

For each solution, clearly indicate the name and the amount of EACH solute AND the amount of solvent (using proper units gms or mls, unless otherwise indicated) that you would use to prepare the solution. Identify the solvent and solute by name and indicate the amount using proper units (gms or mls), unless otherwise indicated. Assume water is the solvent if not indicated.

1.

f. Prepare 250mls of 250mM NaHCO₃

g. Prepare 45 mls of 1.5% agarose in the solvent, TAE

h. Prepare 5L of buffer with 0.5M NaH₂PO₄ and 0.25M Na₂HPO₄

i. Prepare a 80ml reaction mixture with final concentrations of 600mM estradiol, 300mM NaCl, and 1M PBS buffer.

j. How many millimoles of estradiol are contained within the reaction mixture you have just produced (from 1.i.)?

Answer 1

1. 250mls of 250mM NaHCO3:

Stock liquid NaHCO3 = 1M. So dilute 1/4 th to make 250 mM. 250/4 ml = 62.5 ml of NaHCO3 solution. Rest = (250-62.5)ml = 187.5 ml water.

Thus solute = liquid NaHCO3 = 62.5 ml

Solvent = distilled water = 187.5 ml

2. 45 mls of 1.5% agarose in the solvent, TAE

Amount of agarose needed = (45x1.5)/100 grams = 0.675 grams = 675 miligrams

Thus solute = agarose = 675 miligrams

Solvent = TAE = 45 ml

3. 5L of buffer with 0.5M NaH2PO4 and 0.25M Na2HPO4

Solid NaH2PO4 mol wt = 120. For 0.5M, amount needed = moles x mol wt x vol (in litre) = 0.5 x120 x 5 gram = 300 grams

Solid Na2HPO4 mol wt = 142. For 0.25M, amount needed = moles x mol wt x vol (in L) = 0.25 x142 x 5 grams = 177.5 grams

Thus solute = NaH2PO4 = 300 grams

Solute = Na2HPO4 = 177.5 grams

Solvent = distilled water = 5 L

4. 80ml reaction mixture with final concentrations of 600mM estradiol, 300mM NaCl, and 1M PBS buffer

80 ml = 80/1000 L = 0.08 L

Solid Estradiol mol wt = 272.38, For 600 mM (or 0.6M) amount needed= 0.6 x 272.38 x 0.08 grams = 13.07 grams

Solid NaCl mol wt = 58. For 300mM (or 0.3M) amount needed = 0.3 x 58 x 0.08 grams = 1.4 grams

Stock PBS buffer = 3M, thus it needs to get diluted one third in the final volume to make 1M.

Final volume of reaction mixture= 80 ml. So amount of PBS = 80/3 ml = 26.7 ml. Rest = water = 80-26.7 ml = 53.3 ml

Thus solute = estradiol = 13.07 grams

Solute = NaCl = 1.4 grams

Solute = PBS = 27.7 ml

Solvent = distilled water = 53.3 ml

5. Number of milimoles of estradiol in the final reaction mixture = 600 (since the concentration was itself given in milimolar, so the number of milimoles is found out directly. There are 600 milimoles of estradiol in 80 ml reaction mixture.)