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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.
Wiki User
∙ 10y ago
Wiki User
∙ 7y agoThe 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.
Wiki User
∙ 7y ago1. 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
Blessings Mabhena
Dilute 35.05 ml of ammonia with 964.95ml distilled water
Wiki User
∙ 11y agodissolve 1.0 mole gas in 1.0 litre of water
Wiki User
∙ 9y agoTo 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.
Wiki User
∙ 11y ago2.5 g ammonium formate in 1000mL of water
Wiki User
∙ 10y agohow to prepare 5 mM Ammonium formate
gajanan bansode
How to calculate 5 mm ammonium formate in 500ml
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1 mL of 10 mM NaCl solution was mixed with 4 mL of 0.05 M CaCl2 solution The final concentration of Cl- is?
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1 mL of 10 mM NaCl solution was mixed with 4 mL of 0.05 M CaCl2 solution The final concentration of Cl- is?
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