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How much grams of kmno4 to prepare 1000 ppm solution?
To prepare a 1000 ppm (parts per million) solution of KMnO4 (potassium permanganate), you need 1000 mg of KMnO4 per liter of solution. Since 1 gram equals 1000 mg, you would need 1 gram of KMnO4 dissolved in enough water to make a final volume of 1 liter. Therefore, to prepare a 1000 ppm solution, dissolve 1 gram of KMnO4 in 1 liter of water.
How many grams in 0.2N normality KMnO4 solution?
To find the grams of KMnO4 in a 0.2N solution, you need the equivalent weight of KMnO4, which is approximately 158.04 g/mol divided by 5 (since one mole of KMnO4 provides 5 equivalents for redox reactions). Therefore, the equivalent weight is about 31.61 g/equiv. For a 1-liter solution, 0.2N means there are 0.2 equivalents, which translates to about 6.32 grams of KMnO4 (0.2 eq × 31.61 g/equiv).
How to prepare 100 ppm solution of KMnO4?
I will assume that you will start from the crystals of permanganate: Calculations: M.M. potassium permanganate: 158.04 g/mol mol KMnO4 in 10mL sol'n: 1.5 mol/L x 10 mL x (1 L / 1000 mL) = 0.015 mol grams potassium permanganate: 0.015 mol x 158.04 g/mol = 2.3706 g / 10 mL sol'n Preparation: 1. Weigh out analytically 2.3706g KMnO4 into a 10 mL volumetric flask. 2. Dilute to the mark with dH2O.
How many grams of a drug must be used to prepare 3 liters of a 2 percent solution?
To prepare a 2% solution in 3 liters, you would need 60 grams of the drug. This is calculated by multiplying the volume (3 liters) by the percentage (2%) and converting the result to grams. 3 liters x 2% = 60 grams.
How do you prepare 0.1N Na2S2O3 solution?
To prepare a 0.1N Na2S2O3 solution, dissolve 24.98 grams of Na2S2O3·5H2O (sodium thiosulfate pentahydrate) in distilled water and dilute to 1 liter in a volumetric flask. This will give you a 0.1N (normal) solution of Na2S2O3.
How much grams of kmno4 to prepare 1000 ppm solution?
To prepare a 1000 ppm (parts per million) solution of KMnO4 (potassium permanganate), you need 1000 mg of KMnO4 per liter of solution. Since 1 gram equals 1000 mg, you would need 1 gram of KMnO4 dissolved in enough water to make a final volume of 1 liter. Therefore, to prepare a 1000 ppm solution, dissolve 1 gram of KMnO4 in 1 liter of water.
How many grams in 0.2N normality KMnO4 solution?
To find the grams of KMnO4 in a 0.2N solution, you need the equivalent weight of KMnO4, which is approximately 158.04 g/mol divided by 5 (since one mole of KMnO4 provides 5 equivalents for redox reactions). Therefore, the equivalent weight is about 31.61 g/equiv. For a 1-liter solution, 0.2N means there are 0.2 equivalents, which translates to about 6.32 grams of KMnO4 (0.2 eq × 31.61 g/equiv).
How many grams KMnO4 take in 0.1N KMnO4 solution?
(158 g = 1 mole) --- molar mass of potassium permanganate. You also need to specify the volume to be made. For 1 liter just add 15.8 g in a volumetric flask to make 1000 ml (1 liter) of solution.
How do you Prepare 0.02 N potassium permanganate solution?
To prepare a 0.02 N potassium permanganate solution, you would need to dissolve 1.58 grams of potassium permanganate (KMnO4) in 1 liter of distilled water. This will give you a solution with a molarity of 0.02 N. Remember to wear appropriate personal protective equipment when handling potassium permanganate, as it can be harmful.
How do you prepare 30 percent sucrose solution?
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.
How many grams of sulfosalicylic acid are required to prepare 1 liter of 3 solution?
To prepare a 3% solution of sulfosalicylic acid, you would need 30 grams of sulfosalicylic acid for every 1 liter of solution.
How many grams of potassium permanganate KMnO4 are in 2.55 moles?
The compound potassium permanganate has chemical formula KMnO4 Molecular mass of KMnO4 = 39.1 + 54.9 + 4(16.0) = 158.0 Mass of KMnO4 = amount of KMnO4 x Molecular mass of KMnO4 = 2.55 x 158.0 = 403g
How do you prapare 0.01 m potassium permanganate?
To prepare a 0.01 M solution of potassium permanganate (KMnO4), first calculate the amount needed by using the formula: mass (g) = molarity (mol/L) × volume (L) × molar mass (g/mol). The molar mass of KMnO4 is approximately 158.04 g/mol, so for 1 liter of solution, you would need 1.58 grams of KMnO4. Dissolve this amount in a small volume of distilled water, then transfer the solution to a 1-liter volumetric flask and dilute to the mark with distilled water. Mix thoroughly to ensure complete dissolution.
- How many grams of water should be added to prepare a 2% solution by mass with 16.5g of MgCl….A) 402.5B) 808.5C) 155D) 400E) 82.5?
E) 82.5 grams of water should be added.
How many moles of KMnO4 are present in 75.0 mL of a 0.0950 M solution?
Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Algebraically manipulate, moles of solute = Liters of solution * Molarity Moles KMnO4 = (0.075 Liters)(0.0950 M) = 7.13 X 10 -3 moles KMnO4 ------------------------------------
How to prepare 100 ppm solution of KMnO4?
I will assume that you will start from the crystals of permanganate: Calculations: M.M. potassium permanganate: 158.04 g/mol mol KMnO4 in 10mL sol'n: 1.5 mol/L x 10 mL x (1 L / 1000 mL) = 0.015 mol grams potassium permanganate: 0.015 mol x 158.04 g/mol = 2.3706 g / 10 mL sol'n Preparation: 1. Weigh out analytically 2.3706g KMnO4 into a 10 mL volumetric flask. 2. Dilute to the mark with dH2O.
How many grams of active ingredient will you need to prepare 2.5 gallons of a 45.6 percent solution?
4314.9 grams
