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.
First we look at the equation involving KMnO4 in Redox reaction.
MnO4- + 8H+ + 5e- -------> Mn2+ + 4H2O
Hence 5 electrons are transferred in the reaction. Therefore 1N solution of KMnO4 = M/5 solution of KMnO4 i.e., 0.20M KMnO4 solution. Hence 0.1N KMNO4 solution is equivalent to 0.02M KMnO4 i.e., a solution containing 3.1607g/L of KMnO4
1. Weigh 15,803 g ultrapure KMnO4 dried at 110 0C for 30 min.
2. Transfer KMnO4 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.
The molar mass of potassium permanganate is 158,034; 2 M KMnO4 equals 316,068 g.
1000 mL----------------------------316,068 g
20 mL-------------------------------x
x= 20 x 316,068/1000 = 6,32 g
-------------------------------------------------------------------
Preparation: Dissolve 6,32 g KMnO4 a.r. dried in 20 mL demineralized water, at 20 oC, in a volumetric flask.
there is tow method to prepare 1n kmno4 it's depend on oxidation number. Iknow one of this method you need 31.6 gms in one L D.WATER IF YOU KNOW FURTHER INFORMATION PLEASE TELL ME.
By solving 100 mg of KMnO4 in 1 L of distilled water at 20 0C.
Dissolve 0,1 g of potassium permanganate in 100mL water.
Dissolve 1,58 g KMnO4 for analysis in 1 L distilled water at 20 oC.
Dissolve 31,6 g KMnO4 in 1 kg acidified distilled water.
its 100 ppm
5ml of the 1000 ppm solution in 95ml distilled water gives a 500ppm solution
ppm means the amount of solved matter(mg) in one kg of solution. most of the times we use of this equation: (x)ppm=(x)mg/1Liter
100ppm means, you need 100mg in 1oooml. Thus, for 100ml solution, you will need 10mg of methanol to make 100ppm 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
Make a 1 to 100 dilution of the original 1000 ppm solution. That is take 1 ml and dilute to 100 ml, or take 10 ml and dilute to 1000 ml. This will give you a 10 ppm solution.
No 100 ppm of NaCl = 100 mg NaCl/1 L = 67,87 ppm (67,87 mg/L) of Na
7
You prepare a primary solution then dilute portion of your solution down to the required concentration. For example, you want a 10 ppm salt solution (10 mg/L), you dilute 1 g of salt in 1 L of water you get 1000 ppm salt solution. You take 10 ml of your salt solution (0.01 g salt in 10 g) and add in additional 980 ml of water then you get a 10 ppm weight solution.
Assuming that you are referring to PPM as (parts per million), you can prepare a report on the solution, depending on the different ingredients that are involved and quantity of each substance present.
dilute it 1 in 5000. likely best done with a serial or step dilution
200 ppm solution means 200 mg/liter 200 mg = (100 *0.2)/37 = 0.540 g of 37% formaldehyde Vol of 0.540 g of 37 % formaldehyde= 0.540/1.09 = 0.496 mL of 37 % of formaldehyde Therefore: 0.496 mL/1 liter.
1 ppm is 1 mg per litre so you need to weigh out the amount of zinc sulfate that contains 10mg i.e 0.01g SO4 which is 287.4/32 =8.98125g ZnSO4-7H2O and add to a 1 lt volumetric flask and dilute with water to 1 litre.
add 2.4ml glacial aetic acid in one liter of d.m water
Preparation of PPM level standards of ethanol in waterUse the absolute alcohol to prepare the stock solutions of known concentration say for example: 100 ml of 1 g/L of ethanol solution1 g/L is equal to 1000 mg/L which is 1000 ppm. Otherwise the concentration of the 100 ml stock solution is 1000 ppm.Next step: Perform serial dilution procedure to prepare the required ppm level of ethanol standards. For performing the serial dilution, u have to convert the weight of ethanol into volume by dividing the weight by density of ethanol. Density of ethanol = 789 kg/cu.m or 789 g/L.
its 100 ppm