Question #318992

To analyze the alcohol content of a wine, a chemist needs 1.00 L of an aqueous 0.200 M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 must be weighted out to make this solution?


Describe also the laboratory procedures that the chemist must employ in preparing K2Cr2O7 solution.


Expert's answer

Сm(K2Cr2O7) = n(K2Cr2O7)/V(K2Cr2O7) = m(K2Cr2O7)/(Mr(K2Cr2O7)*V(K2Cr2O7))

m(K2Cr2O7) = Сm(K2Cr2O7)*Mr(K2Cr2O7)*V(K2Cr2O7)

Mr(K2Cr2O7) = 2*39 + 2*52 + 7*16 = 78 + 104 + 112 = 294 g/mol

m(K2Cr2O7) = 0.200 M * 294 g/mol * 1.00 L = 58.8 g


So, adding 58.8 g of potassium dichromate to enough water to make 1L will get a 0.2 M solution


sequence of actions:

  • weigh 58.8 g of K2Cr2O7 on electronic scales
  • pour K2Cr2O7 into a flask and add distilled water to the mark of 1 liter
  • stir with a glass rod
  • the solution is ready

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