Dissolve 1,65 g potassium permanganate for analysis in 1 L demineralized water.
4 x. Take 25ml of 0.2 M solution and dilute to 100ml.
pottassium permangate
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.
The equivalent of potassium dichromate is the molar mass/6: 294,1846/6=49,307666. For a 0,1 N solution: 49,30766/10 = 4,93076 g
4 x. Take 25ml of 0.2 M solution and dilute to 100ml.
pottassium permangate
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.
520 ml of HCl in 480 ml of water=1000ml = 5 N
The equivalent of potassium dichromate is the molar mass/6: 294,1846/6=49,307666. For a 0,1 N solution: 49,30766/10 = 4,93076 g
Dilute the Standard Solution (by definition exactly 1 N) 50 times by taking 20 mL 1.00 N sol'n and adding this up to 1.00 L with water
Mix approx. 12 mL of HCl 30 % in 1 L water.
There is no such thing as K2N. K3N is potassium nitride.
add 5 ml of 37% HCl to 495 ml Water. This is 0.12 N ;)
Take 60 gm NaOH (100%) disolve it in distilled water, and make up to 2.0 liter by distlilled water.the prepared solution is 0.75 N NaoH Solution.
We need 1,27 g iodine.