When making small amounts of solutions, it is often convenient to tare or weigh

Question

When making small amounts of solutions, it is often convenient to tare or weigh an empty tube, add a “pinch” of solid reagent, re-weigh the tube, and calculate the volume of solvent in which to dissolve the material. Note that when starting with solid material rather than a solution, the approaches to dilutions that we’ve studied (V1C1 = V2C2; determining DF) do not apply!
1. If you add 0.1431 g of PMSF (phenylmethylsulfonyl fluoride, a protease inhibitor) to a tared tube (so you have an idea of scale, the tube can hold up to 50 mL of solution), what volume of isopropanol should you add to yield a 40 mM final concentration? The formula weight of PMSF is 174.2.
2. What volume of ddH2O should you add to a 1.5 mL microcentrifuge tube (as above, the size is given so you have an idea of scale) to yield a 10% (w/v) solution of ammonium persulfate (APS, the initiator of acrylamide polymerization) given the following:

Mass of empty tube = 1.0692 g.

Mass of tube + APS = 1.1365 g.

Explanation / Answer

1)

we know that

moles = mass / molar mass

so

moles of PMSF = 0.1431 / 174.2

moles of PMSF = 8.2 x 10-4

now

we kow that

conc = moles / volume (L)

here

volume of solution is nothing but volume of isopropanal

so

40 x 10-3 = 8.2 x 10-4 / volume

volume = 8.2 x 10-4 / 40 x 10-3

volume = 0.02053 L

volume = 20.53 ml

so

20.53 ml of isopropanal can be added

2)

Mass of APS = ( mass of tube + APS) - mass of empty tube

= 1.1365 - 1.0692

= 0.0673 g

mass of APS = 0.0673 g

now

let the volume of water be v ml

we know that

1 ml of water = 1 g

so

mass of water = v g

mass of solution = v + 0.0673 g

now

we know that

mass % = (mass of solute / mass of solution ) x 100

so

10 = ( 0.0673 / mass of solution ) x 100

mass of solution = 1.486 g

v + 0.0673 g = 1.486

v = 1.4187

so

1.4187 ml of water should be added

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