Answer to Question #318992 in General Chemistry for Kev

Question

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.

Answer 1

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

m(K₂Cr₂O₇) = С_m(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