1N of k2cr2o7=49.09 gm/1 L
0.25N=12.2725 gm of k2cr2o7 in 1 L of water.
The titer volume of the sample gives the volume of Ferrous Ammonium Sulphate required to react with the excess potassium dichromate in the solution. Similarly, the titer volume for the blank (distilled water) gives the volume of Ferrous Ammonium Sulphate required to react with the excess potassium dichromate in the blank. The equation for the titration can be expressed as: Cr2O72 -- + FeSO4 (NH4)2SO4 = Cr+ + NH4+ + Fe 3+ From above equation it can be seen that one molecule of dichromate corresponds to one molecule of Mohr's salt. Thus, the difference in volume of excess K2Cr2O7 reacting with Mohr's solution can be calculated from the expression: = (Original vol. K2Cr2O7 -- vol. of K2Cr2O7 used for oxidation) solution - (Original vol. K2Cr2O7 -- vol. of K2Cr2O7 used for oxidation) blank = (Vol. of K2Cr2O7 used for oxidation) blank - Vol. of K2Cr2O7 used for oxidation) solution Hence, the difference in the titer volume for the solution and the blank is used to find out the Chemical Oxygen Demand directly.
The chemical formula of potassium dichromate is K2Cr2O7
The chemical forrmula of potassium heptaoxodichromate(VI) is K2Cr2O7. The systematic IUPAC name is potassium dichromate (VI).
When you are looking to make a potassium thicyanate solution you have to dissolve 0.972 g of KSCN in enough H2O to make a 100 mL solution.
K2Cr2O7 + Na2C2O4 --> K2C2O4 + Na2Cr2O7
Weigh 4.9035g K2Cr2O7 make up to 1 liter
The titer volume of the sample gives the volume of Ferrous Ammonium Sulphate required to react with the excess potassium dichromate in the solution. Similarly, the titer volume for the blank (distilled water) gives the volume of Ferrous Ammonium Sulphate required to react with the excess potassium dichromate in the blank. The equation for the titration can be expressed as: Cr2O72 -- + FeSO4 (NH4)2SO4 = Cr+ + NH4+ + Fe 3+ From above equation it can be seen that one molecule of dichromate corresponds to one molecule of Mohr's salt. Thus, the difference in volume of excess K2Cr2O7 reacting with Mohr's solution can be calculated from the expression: = (Original vol. K2Cr2O7 -- vol. of K2Cr2O7 used for oxidation) solution - (Original vol. K2Cr2O7 -- vol. of K2Cr2O7 used for oxidation) blank = (Vol. of K2Cr2O7 used for oxidation) blank - Vol. of K2Cr2O7 used for oxidation) solution Hence, the difference in the titer volume for the solution and the blank is used to find out the Chemical Oxygen Demand directly.
Dissolve 294,185 g of potassium dichromate in 1L demineralized water.
The molarity is 0.001255. Should you really be asking an AP Chem question on Wiki Answers, anyways?
potassium dichromate
What volume of 0.1125 M K2Cr2O7 would be required to oxidize 48.16 mL of 0.1006 M Na2SO3 in acidic solution? The products include Cr3+ and SO42- ions.
K2Cr2O7(aq) ------> 2K+(aq)+Cr2O72-(aq)
The chemical formula of potassium dichromate is K2Cr2O7
Potassium dichromate contain potassium, chromium and oxygen.
potassium dichromate- K2Cr2O7 12.5 grams K2Cr2O7 *(1 mol K2Cr2O7/294 grams K2Cr2O7)= .0425 mols K2Cr2O7 There's two mols of K (potassium) for every 1 mol of K2Cr2O7 (Potassium Dichromate) so you multiply the K2Cr2O7 by two to get mols of K .0425 mols K2Cr2O7*(2 mols K/1 mol K2Cr2O7)=.085 mols K Multiply by the molar mass of K to get grams .85 mols K*(39 grams/1 mol K)= 3.32 grams potassium
No, It's totally different. 1M of K2Cr2O7 is 294.19g/water 1L on the other hands, 1N of K2Cr2O7 is 49.04 g/ water 1L
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