Describe the preparation of 4.00 L of 0.290 M KMnO4 from the solid reagent. ____

ID: 511029 • Letter: D

Question

Describe the preparation of

4.00 L of 0.290 M KMnO4 from the solid reagent.

_____ g KMnO4

4.00 L of 0.290 M HClO4, starting with a 9.00 M solution of the reagent.

_____ mL of 9.00 M HClO4

350 mL of a solution that is 0.0520 M in I-, starting with MgI2.

_____ g MgI2

350 mL of 1.30% (w/v) aqueous CuSO4 from a 0.287 M CuSO4 solution.

_____ mL of the 0.287 M solution

2.50 L of 0.222 M NaOH from the concentrated commercial reagent [50% (w/w), sp gr 1.525].

_____ mL of the concentrated reagent

4.00 L of a solution that is 29.0 ppm in K+, starting with solid K4Fe(CN)6.

_____ mg K4Fe(CN)6

4.00 L of 0.290 M KMnO4 from the solid reagent.

_____ g KMnO4

moles of KMnO4 = Molarity x volume

= 0.290 x 4

= 1.16

mass of KMnO4 = 1.16 x 158 = 183.3 g

mass of KMnO4 = 183.3 g

M1 V1 = M2 V2

4 x 0.290 = 9 x V

V = 0.129L

volume of HClO4 = 129 mL

350 mL of a solution that is 0.0520 M in I-, starting with MgI2.

moles of MgI2 = 2 x 0.0520 x 0.350 = 0.0364

mass of MgI2 = 10.12 g

350 mL of 1.30% (w/v) aqueous CuSO4 from a 0.287 M CuSO4 solution.

1.30 g in 100 mL

? g in 350 mL

mass = 4.55 g

moles = 4.55 / 159.6 = 0.0285

0.0285 = 0.287 x V

V = 99.3 mL

volume = 99.3 mL

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