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Just take 3 grams and add this to 1000 litres (= 1,000 kg = 1,000,000 (million) grams)
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.
.005 grams
1000 grams of water and 2 grams of sugar - sucrose
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)
700gram = 0.7 kilogram
(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.
Just take 3 grams and add this to 1000 litres (= 1,000 kg = 1,000,000 (million) grams)
how to prepare 1000 ppm solution of nickel carbonate
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.
28 gram. = 2 * 56 * 250 / 1000
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.
take 276 gm of salicylic acid in 1000 ml water to prepare 2M solution of the salicylic acid.
.005 grams
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.
for prepare 1mM DPPH radical solution: DPPH Molecular weight: C18H12N5O6 = 394g DPPH 1mM = 394/1000= 0.394g/L
1000 grams of water and 2 grams of sugar - sucrose