Ammonium formate has molecular weight of 63.05
To make 1 L of 1 M, you would weigh 63.05 g and dilute to 1000 mL of water.
The make less, you weigh less e.g. if you want 0.5 M or 500 mM, you wiegh half of 63.05 g and that is 31.52 g and dilute to 1000 mL of water.
The actual procedure will depend on the volume of the solution desired. The following is for 1 liter.
molar mass ammonium formate = 63 g/mole and 100 mM = 100 mmoles/liter = 0.1 mole/liter
0.1 moles/liter x 1 liter x 63 g/mole = 6.3 grams/liter. So, you would dissolve 6.3 grams of ammonium formate in sufficient water to make the FINAL volume 1 liter (1000 mls). To make 100 mls of solution, simply divide by 10 and dissolve 0.63 grams ammonium formate in sufficient water to make a final volume or 100 mls.
1. Weigh 6,306 g ultrapure and dry ammonium formate.
2. Transfer ammonium formate in a clean 1 L volumetric flask using a funnel.
3. Wash the funnel with 0,9 L demineralized water.
4. Put the flask in a thermostat and maintain 30 min at 20 0C.
5. Add demineralized water up to the mark.
6. Stir vigorously and transfer in a clean bottle with stopper.
7. Add a label with necessary information
Dilute 35.05 ml of ammonia with 964.95ml distilled water
dissolve 1.0 mole gas in 1.0 litre of water
To make a 5mM solution of ammonium format, the preparer should mix 10mL of a 9.75% solution of formic acid and water. After this, dilute with water to 15 mL.
2.5 g ammonium formate in 1000mL of water
how to prepare 5 mM Ammonium formate
How to calculate 5 mm ammonium formate in 500ml
[1 mL*10 mM + 4 mL*(50*2 mM Cl-) ] mmolCl- in [1+4] mL total solutionThis is 410 (mmol Cl-) / 5 (mL) = 82 mM Cl-
pl. suggest how to prepare 50 sulphuric acid
for prepare 1mM DPPH radical solution: DPPH Molecular weight: C18H12N5O6 = 394g DPPH 1mM = 394/1000= 0.394g/L
There are five steps on how to make Borax out of Boric Acid. Some of the step-by-step instructions are dilute 25 mm of Hydrochloric Acid with 75 mm of distilled water, put 7g of Borax into a beaker, and then pour in 20 mm of boiling water.
10 mM tartaric acid (sodium) buffer solution (pH=4.2) Tartaric acid (M.W.=150.09)..........................2.5 mmol (0.375 g) Sodium tart rate dihydrate (M.W.=230.08)........7.5 mmol (1.726 g) Add water to make up to 1 L. 10 mM tartaric acid (sodium) buffer solution (pH=2.9) Tartaric acid (M.W.=150.09)..........................7.5 mmol (1.13 g) Sodium tartrate dihydrate (M.W.=230.08)........2.5 mmol (0.58 g) Add water to make up to 1 L.
To prepare the buffer using solid form reagents, prepare a 0.1 M ammonium acetate solution by dissolving 7.7 g ammonium acetate in a 1000 ml water. Adjust 1 L of this solution to pH 4.5 by adding acetic acid (about 8 ml) and 5 ml of 1 M p-TSA (equivalent to 5 mM p-TSA).
I don't know how to make the solution below. Low salt buffer: 10 mM phosphate buffer, 10 mM NaCl, pH 7.4. Could you tell me the method in detail?
Assuming that "mM" means "millimolar", the solution specified contains 6 millimoles of ammonium sulphate per liter. Therefore, 25 ml of the solution contains 6(25/1000) = 0.15 millimoles. By definition, there are 1000 micromoles per millimole. Therefore, 0.15 millimoles = 150 micromoles.
show solution convert 0.015 km to mm
Molecular formulaC2H3O2NH4Molar mass77.0825 g/mol 77.0825 g/mol x .005 molar x 3 liters = 1.1562375 g of Ammonium Acetate dissolved in 3 liters --> 5 mM
Also 150 mM of sodium.
Dilute a measured volume of the 100 mM solution with 19 times its own volume of pure water to produce a 5mM solution.
10-4
FMOT @ Shee_Coldd
[1 mL*10 mM + 4 mL*(50*2 mM Cl-) ] mmolCl- in [1+4] mL total solutionThis is 410 (mmol Cl-) / 5 (mL) = 82 mM Cl-
1000 mm
Multiply mm by 10 to get cm.