From http://en.wikipedia..../Methylene_blue Molar mass=319.85 g/mol
2.303%=2.303g solute / 100mL solution = 2.303g / .1L = (23.03g / 1L) * (1mol/319.85g) = 2303/3985 moles per liter of solution = initial molarity = m1
According to http://www.novalek.c...easurements.htm 1 drop=0.05ml
Therefore for a 60mcg solution,
60mcg/.05mL = 6.0*10^-5g / 5.0*10^-5 L = [(6/5g)/L]*[1mol/319.85g] = 24/6397 moles per liter of solution = final molarity = m2
The dillution equation is m1v1=m2v2 where m1 is the initial molarity, m2 is the final molarity, v1 is the initial volume and v2 is the final volume. M1 and M2 are constant, so you can calculate the final volume given an initial volume.
For example 1/4 teaspoon of 2.303% MB = 0.0012322304L
The final volume = v2 = m1v1/m2 = (2303/31985)(.0012322304L)/(24/6397) = 0.02364855509 L
The amount of water needed is 0.02364855509 L - 0.0012322304L = 0.02241632469L ~ 22.4ml
In summary,
1/4 teaspoon Methylene Blue mixed with 22.4ml water = 1 drop containing 60mcg solution.
Its after 3am here... didn't look at the numbers but assuming all correct then your summary seems to be misstated (in the sense that you seem to be saying all that final solution only yields 1 drop). When from your numbers it appears that the final volume = 0.02364855509 L = ~ 23.6mL and with ~20 metric drops/mL you should have: ~472 drops @ 60mcg/drop
For simplicity and if 1metric drop = 0.05mL = 0.00005L ... Then using the rest of your numbers:
v1 = 1 metric drop of 2.303% (w/v) MB = 0.00005L
The final volume = v2 = m1v1/m2 = (2303/31985)(0.00005L)/(24/6397) = 9.5958333333333333333333333341607e-4L = ~0.96mL
The amount of water needed is 9.5958333333333333333333333341607e-4L - 0.00005L = 9.0958333333333333333333333341607e-4 L = ~0.91mL
So adding 1 drop of 2.303% (w/v) MB to 0.91mL water should give us ~0.96ml of diluted solution.
and with ~20 metric drops/mL you should have: ~ 19 drops @ 60mcg/drop
Those are real small numbers and it doesn't give us much final solution. So using 1mL for initial volume and your other numbers....
v1 = 1mL of our 2.303% (w/v) MB = 0.001L
The final volume = v2 = m1v1/m2 = (2303/31985)(0.001L)/(24/6397) = 0.019191666666666666666666666668422L = ~19.2mL
The amount of water needed is 0.019191666666666666666666666668422L - 0.001L = 0.018191666666666666666666666668L = ~18.2mL
So adding 1mL of 2.303% (w/v) MB to 18.2mL of water should give us ~19.2ml of diluted solution.
and with ~20 metric drops/mL you should get: ~ 384 drops @ 60mcg/drop
~~~~~~~~~~~~~~
If you notice the posts before this one, even the simple calculations were pretty close. Without knowing the Molar mass of MB or using these formulas, they would have arrived at basically the same answer given a 2.303% (w/v) solution has:
2.303 g solute in 100mL solution
or 0.02303g in 1mL
or 23.03mg in 1mL
or 23030mcg in 1mL
20 metric drops in 1mL (0.05mL/drop) - see
http://en.wikipedia....i/Drop_(volume)23030mcg / 20 drops = 1151.5mcg/drop
So there is ~ 1151.5mcg MB per metric drop of a 2.303% (w/v) MB solution
but we want 60mcg/drop. Our concentration is 19.1917 (1151.5 / 60) times too strong.
So going back to mL to recheck ... 60mcg/drop ~= 1200mcg/mL
(23030 / 1200) = 19.191666666666666666666666666667
Our 2.303% solution is 23030mcg in 1mL... but we need for it to be 23030mcg in ~19.2mL. So we add ~18.2mL of water to our existing 1mL of 2.303% solution.
and with ~20 metric drops/mL you should get: ~ 384 drops @ 60mcg/drop
edit:fixed line wraps
Edited by frankbuzin, 19 October 2009 - 07:45 AM.