what is the required weight( in milligrams) to prepare 6.0m of potassium dichromate with a total volume of 4 liters.
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what is the required weight( in milligrams) to prepare 6.0m of potassium dichromate with a total volume of 4 liters.
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Answer:
7060000 mg of Potassium Dichromate
Explanation:
To Compute the required mass of the substance or solute
M = mol/L
6.0 M = mol/1
mol = 6
6 mol / 1 x 4L = 24 mol of Potassium Dichromate
To convert in grams simply multiply the mol to molar mass of the solute
In this case Potassium Dichromate has a molar mass of 294.185g/mol
24mol x 294.185g/mol * Cancel mol
= 7060.44 g of Potassium Dichromate
To Convert in mg simply multiply to 1000
7060.44g of Potassium Dichromate x 1000 = 7060000 mg of Potassium Dichromate