Suppose 300 is mol wt of compound
300g in 1000 ml -- it becomes 1M
300,000 mg in 1000 ml ---it is also 1 Molar
1 mg=1000 microgram
300,000,000 microgram in 1000 ml ----it is 1 molar
300,000 microgram in 1 ml ----it is 1 Molar
1 molar=1000 milimolar
300,000 microgram in 1 ml -----1000 milimolar
300,000 ----------------------------- 1000,000 micromolar
0.3 microgram --in 1 ml it is 1 micromolar
simillarly convert the ml as you want
smiley by dividing molar solution by 1000
take 1ml from stock solution and dilute to 10ml of water to get 10 micromolar solution
Dissolve 2ml from stock in 8ml of diluent.
The easiest way would be to take a 25 ml aliquot of the 50 micromolar solution with a pipette and transfer to a 50 ml volumetric flask and then make up to the mark with distilled water.
Maltose and water react to form the maltose solution. A sweet solution!
dilute it twofold. this means if you choose a 100ml volumetric, take 50ml of your stock and fill it to the mark (another 50ml) with solvent
The meaning of micromolar is 1/1 000 000 mol.
5 millimolar (5 thousandths of a mole per liter) is equal to 5000 micromolar ( 5000 millionths of a mole per liter). To make a 50 micromolar solution from 5 millimolar stock solution, you therefore need 5000/50 = 100 fold dilution. Remove a 10 ml aliquot of stock and transfer to a 1000ml (1 liter) volumetric flask. Dilute with the solvent -usually water, and fill up to the graduation mark. You will now have 1 liter (1000 ml) of 50 micromolar solution.
Maltose and Lactose are "Reducing Sugar's" in solution with Benedict's Reagent, while sucrose is not.
Any dilution of 100,000 (= 1,000,000 / 10) will do, eg. 1 mL added up to 100 L.
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There are various characteristics of maltose. Some of them include the ability to reduce the Fehling solution. This is as result of having free aldehyde.
Maltose. Water and Starch mixed with amylase makes maltose
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See the two Related Questions to the left for the answer.The first is how to prepare a solution starting with a solid substance (and dissolving it). The second question is how to prepare a solution by diluting another solution.
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- Divide the molar mass of the compound with 1 000 000. - Multiply with 200. - Weigh this mass if it is possible; alternatively obtain this mass from a stock solution. - Dissolve this compound in 1 L of demineralized water.