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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.
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How do you prepare 3 ppm solution of ferric chloride in 1000 liter water?
Just take 3 grams and add this to 1000 litres (= 1,000 kg = 1,000,000 (million) grams)
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.
What is the total mass of solute in 1000 grams of solution having a concentration on 5 ppm?
5 ppm means that there are 5 grams of solute in every 1 million grams of solution. Therefore, in a 1000-gram solution, the total mass of solute would be 0.005 grams.
What are the components of sugar solution?
1000 grams of water and 2 grams of sugar - sucrose
How do you convert grams into millograms?
You need to know the volume of liquid that contains the solid to answer this question. If you have one liter of a solution that contains 0.1 grams per milliliter, you can find the number of grams in the container through this calculation: 1 liter X 1000 ml/liter X 0.1 grams/ml = 100 grams of solid in the one liter of solution (the units for liters cancel out)
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 100 ml of 1 in 1000 KMnO4?
700gram = 0.7 kilogram
How do you prepare 3 ppm solution of ferric chloride in 1000 liter water?
Just take 3 grams and add this to 1000 litres (= 1,000 kg = 1,000,000 (million) grams)
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 do you prepare 2 ppm solution of nickel nitrate?
To prepare a 2 ppm solution of nickel nitrate, you would dissolve 2 grams of nickel nitrate in enough water to make 1 liter of solution. This will result in a solution where there are 2 parts of nickel nitrate for every 1 million parts of water.
How do you prepare the 1N concentration of potassium permamganate solution?
N (normality) describes a solution that contains 1 gram equivalent weight (gEW) per liter solution. An equivalent weight is equal to the molecular weight divided by the valence (here it gets a little tricky, for acids ands bases it refers to the number of H+ or OH-, in salts it must be expressed which ion is meant unless the ratio is 1:1). In the case of KMnO4, equivalent wt is reaction specific. When KMnO4 is used in acid medium as oxidiser, 5 electrons are gained by Mn atom. So equivalent wt of KMnO4 in acid medium = Molecular wt/no.of electrons gained in redox reaction = 158/5 =31.6. So for 0.1N KMnO4 solution, you have to dissolve 3.16g KMnO4 in 1L water. (Usually a little bit excess is taken, say 3.25g, since some crystals of KMnO4 will be remained undissolved that have to be removed by filtration. So eventhough u r preparing 0.1N KMnO4 solution by accurate weighing,it is not a primary standard and u have to standardise it against a primary std such as oxalic acid or sodium oxalate. In alkaline or neutral medium, reaction of KMnO4 is different and Mn gains 3 electrons in redox reaction. So, for alkaline medium redox titrations, equivalent wt of KMnO4 will be 158/3 = 52.6. So for 0.1N KMnO4 solution in alkaline medium redox titration, dissolve 5.26g in 1L water.
How many grams of dextrose will 1000 ml of a 10 percent solution contain?
A 10 percent solution of dextrose means that there are 10 grams of dextrose per 100 milliliters of solution. Therefore, in 1000 milliliters (which is 10 times 100 ml), there would be 10 grams x 10 = 100 grams of dextrose in a 1000 ml solution.
What is the total mass of solute in 1000 grams of solution having a concentration on 5 ppm?
5 ppm means that there are 5 grams of solute in every 1 million grams of solution. Therefore, in a 1000-gram solution, the total mass of solute would be 0.005 grams.
How do you prepare 0.01n sodium thiosulfate as per ip?
To prepare 0.01N sodium thiosulfate per Indian Pharmacopoeia (IP) standards, dissolve 25.3 grams of pure sodium thiosulfate pentahydrate (Na2S2O3·5H2O) in distilled water to make 1000 ml of solution. This solution will have a normality of 0.01N.
If you have 17.34 moles of potassium permanganate how many kilograms of the substance do you have?
Potassium permanganate = KMnO4 Molar mass of KMnO4 K = 1 * 39.10 g = 39.10 g Mn = 1 * 54.94 g = 54.94 g O = 4 * 16.00 g = 64.00 g Total = 158.04 g/mol 17.34 mol KMnO4 * (158.04 g/1 mol KMnO4) = 2740.41 g KMnO4 Convert the grams to kilograms. 1 kg = 1000 g 2740.41 g * (1 kg/1000 g) = 2.74041 kg Therefore, 17.34 moles of potassium permanganate is equal to about 2.74 kilograms.
How can you prepare solution 0.2 ppm from 1000 ppm?
To prepare a 0.2 ppm solution from a 1000 ppm solution, you would need to dilute the 1000 ppm solution by adding 5000 parts of solvent for every 1 part of the 1000 ppm solution. This means mixing 1 part of the 1000 ppm solution with 5000 parts of solvent to achieve a 0.2 ppm concentration.
How many grams of salt are in an 80 parts per thousand liter solution of salt water?
80 parts per thousand (ppt) means there is 80 grams (g) of NaCl (salt) per 1000 grams of water. 1L is approximately 1000 grams of water; it can vary slightly depending on the temperature.
