1/5 of molecular weight
for preparing 0.1 normal solution of potassium permanganate you have to disssolve 3.16 g potssium permangnate in 1L water bt in alkaline or neutral medium reactions of potassium permanganate is different and Mn gains 3 electrons in redox reaction,so far alkaline medium redox titration equivalent wt of KMnO4 will be 158\3=52.6.so far,0.1 N KMnO4 in alkaline medium redox titration dissolve 5.26 g in 1L sol.
This is because in acidic medium kmn04 decomposes to give Mn2+ ions which impart pink colour to the solution. The reaction being Mno4- +8H+5e ----->Mn2+ + 4H2O
31.608 g/mol
An alkaline medium is formed when alkaline elements in food break down. This happens when they combine with (H+) ions.
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.
for preparing 0.1 normal solution of potassium permanganate you have to disssolve 3.16 g potssium permangnate in 1L water bt in alkaline or neutral medium reactions of potassium permanganate is different and Mn gains 3 electrons in redox reaction,so far alkaline medium redox titration equivalent wt of KMnO4 will be 158\3=52.6.so far,0.1 N KMnO4 in alkaline medium redox titration dissolve 5.26 g in 1L sol.
Potassium permanganate dye diffuses more quickly through water than agar gel.The rate of diffusion depends on the molecular weight of the chemical and the characteristics of the medium through which the substance diffuses.
Potassium permanganate dye diffuses more rapidly through water than agar gel. The rate of diffusion depends on the molecular weight of the chemical and the characteristics of the medium through which the substance diffuses.
This is because in acidic medium kmn04 decomposes to give Mn2+ ions which impart pink colour to the solution. The reaction being Mno4- +8H+5e ----->Mn2+ + 4H2O
31.608 g/mol
An alkaline medium is formed when alkaline elements in food break down. This happens when they combine with (H+) ions.
simply because the chemical gives a very dark/strong purple color. You'll be able to see where the convection currents take the chemical in stagnant water when the medium (water) is heated...
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.
Ginger root is medium-level alkaline
the color of methyl orange in an acidic medium is orange
1 medium-sized fig has about 116mg of potassium.
strong alkali