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How much gm of disodium edta require to prepare 0.05 m disodium edta solution?
To prepare a 0.05 M disodium EDTA solution, you would need to dissolve 3.72 grams of disodium EDTA dihydrate (Na2C10H14N2Na2·2H2O) in enough water to make 1 liter of solution.
How much gm of disodium EDTA to prepare O5 M disodium EDTA?
To prepare a 0.5 M solution of disodium EDTA, you would need to calculate the molar mass of disodium EDTA (approximately 372.24 g/mol) and then use the formula: (given concentration x volume in liters) / molar mass = mass needed. So, for 0.5 M solution: (0.5 mol/L x 1 L) / 372.24 g/mol = 0.00134 kg or 1.34 g of disodium EDTA is needed.
How do you prepare M Cal solution?
To prepare 1 M CaI aqueous solution, dissolve 29.4 g in a total volume of 100 mls, or 294 g in a total volume of 1 liter.
How would one prepare 10 ml of a 25 M HCI solution if 1 M HCI was available?
To prepare a 10 ml of a 25 M HCl solution, you would need to dilute the 1 M HCl solution by a factor of 25. You will need 0.25 ml of the 1 M HCl solution and 9.75 ml of solvent (usually water) to make a total volume of 10 ml for the 25 M HCl solution.
What is structure of EDTA?
according to http://www.sciencelab.com/ all these.....Dihydrogen Magnesium Ethylenediaminetetraacetate » Edetate Calcium Disodium, FCC» Edetate Disodium Dihydrogen» Edetate Disodium TS» Edetate Disodium, Dihydrate, FCC» Edetate Disodium, Dihydrate, Reagent, ACS» Edetate Disodium, Dihydrate, USP» Edetic Acid, NF» EDTA Disodium, 0.01 M Solution» EDTA Disodium, 0.01 M Solution, w/Magnesium» EDTA Disodium, 0.05 M Solution» EDTA Disodium, 0.0575 M Solution» EDTA Disodium, 0.1 M Solution» EDTA Disodium, 0.5 M Solution» EDTA Disodium, 10% (w/v) Solution» EDTA Disodium, 2.5% (w/v) Aqueous Solution» EDTA Stabilizer Solution, 500 g/L» Ethylenediaminetetraacetate Acid Diammonium Dihydrogen Salt» Ethylenediaminetetraacetic Acid Calcium Disodium Salt» Ethylenediaminetetraacetic Acid Copper (II) Disodium Salt» Ethylenediaminetetraacetic Acid Dicalcium Salt» Ethylenediaminetetraacetic Acid Dimagnesium Salt» Ethylenediaminetetraacetic Acid Dipotassium Salt, Reagent» Ethylenediaminetetraacetic Acid Iron (III) Ammonium Salt, Solution» Ethylenediaminetetraacetic Acid Magnesium Disodium Salt» Ethylenediaminetetraacetic Acid Tetrasodium Salt, Reagent» Ethylenediaminetetraacetic Acid Tetrasodium Salt, Solution» Ethylenediaminetetraacetic Acid Tripotassium Salt, Dihydrate» Ethylenediaminetetraacetic Acid Trisodium Salt» Ethylenediaminetetraacetic Acid, Dipotassium Magnesium Salt, Dihydrate, USP» Ethylenediaminetetraacetic Acid, Iron (III) Sodium Salt» Ethylenediaminetetraacetic Acid, Powder, Reagent, ACS» Magnesium EDTA, 0.02% (w/v) Solutionhope this was helpful
How much gm of disodium edta require to prepare 0.05 m disodium edta solution?
To prepare a 0.05 M disodium EDTA solution, you would need to dissolve 3.72 grams of disodium EDTA dihydrate (Na2C10H14N2Na2·2H2O) in enough water to make 1 liter of solution.
How much quantity of EDTA dissolved in 1000ml to get 0.01 m solution of EDTA?
To prepare a 0.01 M solution of EDTA in 1000 ml, you would need 37.22 grams of EDTA disodium salt dihydrate (C10H14N2Na2O8·2H2O) or approximately 0.1 moles. Dissolve the EDTA in water and make up the volume to 1000 ml to get a 0.01 M solution.
How much gm of disodium EDTA to prepare O5 M disodium EDTA?
To prepare a 0.5 M solution of disodium EDTA, you would need to calculate the molar mass of disodium EDTA (approximately 372.24 g/mol) and then use the formula: (given concentration x volume in liters) / molar mass = mass needed. So, for 0.5 M solution: (0.5 mol/L x 1 L) / 372.24 g/mol = 0.00134 kg or 1.34 g of disodium EDTA is needed.
How do you can prepare 0.5molar EDTA solution?
To prepare a 0.5 M EDTA solution, dissolve the appropriate amount of EDTA disodium salt dihydrate (molecular weight 372.24 g/mol) in water to achieve a final volume desired. For example, to make 100 mL of 0.5 M EDTA solution, you would dissolve 18.61 g of EDTA disodium salt dihydrate in water and adjust the volume to 100 mL.
How do you prepare 5mM EDTA of 100ml solution?
5mM = 0.005 moles 100 mL = 0.1 Liters Molarity = moles of solute/Liters of solution 0.005 M EDTA = X moles/0.1 Liters = 0.0005 moles EDTA =_____________ Now, look up the molecular formula for EDTA and find how many grams needed to add to your 100 mL.
How do you prepare M Cal solution?
To prepare 1 M CaI aqueous solution, dissolve 29.4 g in a total volume of 100 mls, or 294 g in a total volume of 1 liter.
A corrosion technologist pipetted a 100 ml hard water sample and titrated it with 37.64 ml of 0.01 m edta solution for a total hardness endpoint and 29.32 ml of 0.01 m edta solution for a calcium?
The total hardness of the water sample is calculated as 37.64 ml of 0.01 M EDTA solution. The calcium hardness is calculated as the difference between the total hardness and the magnesium hardness, which is 29.32 ml of 0.01 M EDTA solution. It's important to conduct proper titration techniques to ensure accurate results for water quality assessment.
How do you prepare 1 M solution from 6 M solution?
It's simple to dilute a solution, just add more water. If you have one liter of 6 molar solution, and add another 5 liters of water, it becomes a 1 molar solution. You can also use M1V1 = M2V2 formula for dilution. For example, if you want to prepare 1 M of solution in 1 litre of water, then how much volume you need from the initial 6 M solution? Simply use the equation, (6 M) x (V1) = (1 M) x (1 litre) ---> V1 = (1 M) x (1 litre)/(6 M) = 0.167 L or 167 ml
How do you prepare 1 L of 15 M sodium hydroxide solution by diluting the stock solution of 6 M sodium hydroxide?
To prepare 1 L of 15 M NaOH solution, you need to dilute the 6 M stock solution. You would need to take 500 mL of the 6 M NaOH stock solution and add water to bring the total volume up to 1 L. This will give you a final concentration of 15 M.
How would one prepare 10 ml of a 25 M HCI solution if 1 M HCI was available?
To prepare a 10 ml of a 25 M HCl solution, you would need to dilute the 1 M HCl solution by a factor of 25. You will need 0.25 ml of the 1 M HCl solution and 9.75 ml of solvent (usually water) to make a total volume of 10 ml for the 25 M HCl solution.
What is structure of EDTA?
according to http://www.sciencelab.com/ all these.....Dihydrogen Magnesium Ethylenediaminetetraacetate » Edetate Calcium Disodium, FCC» Edetate Disodium Dihydrogen» Edetate Disodium TS» Edetate Disodium, Dihydrate, FCC» Edetate Disodium, Dihydrate, Reagent, ACS» Edetate Disodium, Dihydrate, USP» Edetic Acid, NF» EDTA Disodium, 0.01 M Solution» EDTA Disodium, 0.01 M Solution, w/Magnesium» EDTA Disodium, 0.05 M Solution» EDTA Disodium, 0.0575 M Solution» EDTA Disodium, 0.1 M Solution» EDTA Disodium, 0.5 M Solution» EDTA Disodium, 10% (w/v) Solution» EDTA Disodium, 2.5% (w/v) Aqueous Solution» EDTA Stabilizer Solution, 500 g/L» Ethylenediaminetetraacetate Acid Diammonium Dihydrogen Salt» Ethylenediaminetetraacetic Acid Calcium Disodium Salt» Ethylenediaminetetraacetic Acid Copper (II) Disodium Salt» Ethylenediaminetetraacetic Acid Dicalcium Salt» Ethylenediaminetetraacetic Acid Dimagnesium Salt» Ethylenediaminetetraacetic Acid Dipotassium Salt, Reagent» Ethylenediaminetetraacetic Acid Iron (III) Ammonium Salt, Solution» Ethylenediaminetetraacetic Acid Magnesium Disodium Salt» Ethylenediaminetetraacetic Acid Tetrasodium Salt, Reagent» Ethylenediaminetetraacetic Acid Tetrasodium Salt, Solution» Ethylenediaminetetraacetic Acid Tripotassium Salt, Dihydrate» Ethylenediaminetetraacetic Acid Trisodium Salt» Ethylenediaminetetraacetic Acid, Dipotassium Magnesium Salt, Dihydrate, USP» Ethylenediaminetetraacetic Acid, Iron (III) Sodium Salt» Ethylenediaminetetraacetic Acid, Powder, Reagent, ACS» Magnesium EDTA, 0.02% (w/v) Solutionhope this was helpful
How do you prepare a solution of 1 M HClO4?
To prepare a 1 M solution of perchloric acid (HClO4), you would need to dissolve 1 mole of HClO4 in enough water to make a final volume of 1 liter. This can be calculated by using the formula: moles = Molarity (M) x Volume (L). Please handle perchloric acid with caution and use appropriate safety measures while preparing the solution.
